Maths-

General

Easy

Question

The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

⁴C₃ × ⁵C₃ ×⁶C₃
⁴P₃×⁵P₃ × ⁶P₃
$\frac{15!}{4! 5! 6!}$
$\frac{15!}{(3!)^{3}}$

Hint:

Here, we can use the formula
$C presuperscript n subscript r space cross times r factorial equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r$

The correct answer is: ⁴P₃×⁵P₃ × ⁶P₃

Given that,
Shirts = 4 , Pants = 5 and Hats = 6, which are distributed among 3 men.
For shirts, Number of ways they can wear = $C presuperscript 4 subscript 3 space end subscript cross times 3 factorial$
For pants, Number of ways they can wear = $C presuperscript 5 subscript 3 cross times 3 factorial$
For hats, number of ways they can wear = $C presuperscript 6 subscript 3 cross times 3 factorial$
Total number of ways they can wear
$rightwards double arrow left parenthesis C presuperscript 4 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 5 subscript 3 cross times 3 factorial right parenthesis cross times left parenthesis C presuperscript 6 subscript 3 cross times 3 factorial right parenthesis rightwards double arrow fraction numerator 4 factorial over denominator 3 factorial thin space 1 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 5 factorial over denominator 3 factorial thin space 2 factorial end fraction cross times 3 factorial cross times thin space fraction numerator 6 factorial over denominator 3 factorial thin space 3 factorial end fraction cross times 3 factorial rightwards double arrow fraction numerator 4 factorial over denominator 1 factorial end fraction cross times fraction numerator 5 factorial over denominator 2 factorial end fraction cross times fraction numerator 6 factorial over denominator 3 factorial end fraction space left parenthesis C presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r right parenthesis rightwards double arrow P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3$

Thus, the number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is $P presuperscript 4 subscript 3 cross times P presuperscript 5 subscript 3 cross times P presuperscript 6 subscript 3$ .

Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-

Maths-General

General

Maths-

Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-

Maths-General

General

Maths-

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

DETAILED SOLUTION:

There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.

Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are

C presuperscript 10 subscript 6

,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are

C presuperscript 4 subscript 4 space end subscript equals space 1

And now all the 16 people are placed in their seats according to the constraints.

Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is

(8! \times 8!)

.
So, a possible number of arrangements will be

rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

Now as we know

C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210 A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

So total number of arrangements is

= 210 \times (8! \times 8!)

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

Maths-General

C presuperscript 10 subscript 6

,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are

C presuperscript 4 subscript 4 space end subscript equals space 1

(8! \times 8!)

.
So, a possible number of arrangements will be

rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

Now as we know

C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210 A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

So total number of arrangements is

= 210 \times (8! \times 8!)

General

Maths-

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

Maths-General

General

physics-

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript

equals negative U subscript P end subscript equals negative m V subscript P end subscript

equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses

Potential at point

P

will be obtained by in integration as given below. Let

d M

be the mass of small rings as shown

d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r

equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction

d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction

equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r

equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses

therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

physics-General

W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript

equals negative U subscript P end subscript equals negative m V subscript P end subscript

equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses

Potential at point

P

will be obtained by in integration as given below. Let

d M

be the mass of small rings as shown

d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r

equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction

d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction

equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r

equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses

therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

General

physics-

The two bodies of mass $m subscript 1 end subscript$ and $m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

In the pulley arrangement

open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g

But

stack a with rightwards arrow on top subscript 1 end subscript

is in downward direction and in the upward direction

i e

stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript

therefore

Acceleration of centre of mass

stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction

equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g

The two bodies of mass $m subscript 1 end subscript$ and $m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

physics-General

In the pulley arrangement

open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g

But

stack a with rightwards arrow on top subscript 1 end subscript

is in downward direction and in the upward direction

i e

stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript

therefore

Acceleration of centre of mass

stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction

equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g

General

maths-

If $x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis.$ to $straight infinity$ terms, $vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis$ $straight infinity$ terms, $vertical line a vertical line less than 1$ , then $x colon y$

maths-General

General

Maths-

The coefficient of $x to the power of negative n end exponent$ in $left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent$ is

Maths-General

General

maths-

$open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis$ $plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p$

maths-General

General

chemistry-

Compounds (A) and (B) are –

chemistry-General

General

Maths-

$2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals$

2. C subscript 0 plus 5. C subscript 1 plus 8 C subscript 2 plus........... plus left parenthesis 3 n plus 4 right parenthesis C subscript n equals sum from r equals 0 to n of left parenthesis 3 n plus 4 right parenthesis space C subscript r equals sum from r equals 0 to n of 3 n. space C subscript r plus sum from r equals 0 to n of 2. C subscript r space space space space space space space space space space space space space space space open square brackets a s space r equals 0 comma space s o comma space space C subscript r equals 0 close square brackets equals sum from r equals 1 to n of 3 n. space C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript plus sum from r equals 0 to n of 2. C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript space space space space space space space space space space space open square brackets a s space 2 to the power of n equals C presuperscript n subscript 0 plus C presuperscript n subscript 1 plus C presuperscript n subscript 2 plus C presuperscript n subscript 3......... plus C presuperscript n subscript n close square brackets equals 3 n.2 to the power of n minus 1 end exponent plus 2.2 to the power of n minus 1 end exponent equals left parenthesis 2 plus 3 n right parenthesis 2 to the power of n minus 1 end exponent space space space space

$2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals$

Maths-General

2. C subscript 0 plus 5. C subscript 1 plus 8 C subscript 2 plus........... plus left parenthesis 3 n plus 4 right parenthesis C subscript n equals sum from r equals 0 to n of left parenthesis 3 n plus 4 right parenthesis space C subscript r equals sum from r equals 0 to n of 3 n. space C subscript r plus sum from r equals 0 to n of 2. C subscript r space space space space space space space space space space space space space space space open square brackets a s space r equals 0 comma space s o comma space space C subscript r equals 0 close square brackets equals sum from r equals 1 to n of 3 n. space C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript plus sum from r equals 0 to n of 2. C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript space space space space space space space space space space space open square brackets a s space 2 to the power of n equals C presuperscript n subscript 0 plus C presuperscript n subscript 1 plus C presuperscript n subscript 2 plus C presuperscript n subscript 3......... plus C presuperscript n subscript n close square brackets equals 3 n.2 to the power of n minus 1 end exponent plus 2.2 to the power of n minus 1 end exponent equals left parenthesis 2 plus 3 n right parenthesis 2 to the power of n minus 1 end exponent space space space space

General

maths-

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

maths-General

General

Maths-

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

Let one of the roots of the equation

a x squared plus b x plus c equals 0

alpha

.
Then as per question one of the roots of the equation

a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0

will be

1 over alpha

a x squared plus b x plus c equals 0 rightwards double arrow a alpha squared plus b alpha plus c equals 0 space _ _ _ _ _ _ _ _ _ e q u a t i o n space 1

a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 rightwards double arrow a subscript 1 open parentheses 1 over alpha close parentheses squared plus b subscript 1 open parentheses 1 over alpha close parentheses plus c subscript 1 equals 0 _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 2

Multiplying equation 1 by

c subscript 1

and equation 2 by a, then subtracting, we get:

stack attributes charalign center stackalign right end attributes row a c subscript 1 alpha squared plus b c subscript 1 alpha plus c c subscript 1 equals 0 end row row a c subscript 1 alpha squared plus a b subscript 1 alpha plus a subscript 1 a equals 0 end row row minus none end row horizontal line row b c subscript 1 alpha minus a b subscript 1 alpha plus c c subscript 1 minus a subscript 1 a equals 0 end row end stack

rightwards double arrow alpha open parentheses b c subscript 1 minus b subscript 1 a close parentheses plus c c subscript 1 minus a subscript 1 a equals 0 rightwards double arrow alpha equals fraction numerator a subscript 1 a minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction

putting the value in equation 1

a open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses squared plus b open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses plus c equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus b open parentheses a a subscript 1 minus c c subscript 1 close parentheses open parentheses b c subscript 1 minus a b subscript 1 close parentheses plus c open parentheses b c subscript 1 minus a b subscript 1 close parentheses squared equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses open parentheses a a subscript 1 b minus c c subscript 1 b plus a b subscript 1 c plus b c c subscript 1 close parentheses equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals negative open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis rightwards double arrow open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus a b subscript 1 close parentheses left parenthesis b subscript 1 c minus a subscript 1 b right parenthesis

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

Maths-General

Let one of the roots of the equation

a x squared plus b x plus c equals 0

alpha

.
Then as per question one of the roots of the equation

a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0

will be

1 over alpha

a x squared plus b x plus c equals 0 rightwards double arrow a alpha squared plus b alpha plus c equals 0 space _ _ _ _ _ _ _ _ _ e q u a t i o n space 1

a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 rightwards double arrow a subscript 1 open parentheses 1 over alpha close parentheses squared plus b subscript 1 open parentheses 1 over alpha close parentheses plus c subscript 1 equals 0 _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 2

Multiplying equation 1 by

c subscript 1

and equation 2 by a, then subtracting, we get:

stack attributes charalign center stackalign right end attributes row a c subscript 1 alpha squared plus b c subscript 1 alpha plus c c subscript 1 equals 0 end row row a c subscript 1 alpha squared plus a b subscript 1 alpha plus a subscript 1 a equals 0 end row row minus none end row horizontal line row b c subscript 1 alpha minus a b subscript 1 alpha plus c c subscript 1 minus a subscript 1 a equals 0 end row end stack

rightwards double arrow alpha open parentheses b c subscript 1 minus b subscript 1 a close parentheses plus c c subscript 1 minus a subscript 1 a equals 0 rightwards double arrow alpha equals fraction numerator a subscript 1 a minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction

putting the value in equation 1

a open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses squared plus b open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses plus c equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus b open parentheses a a subscript 1 minus c c subscript 1 close parentheses open parentheses b c subscript 1 minus a b subscript 1 close parentheses plus c open parentheses b c subscript 1 minus a b subscript 1 close parentheses squared equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses open parentheses a a subscript 1 b minus c c subscript 1 b plus a b subscript 1 c plus b c c subscript 1 close parentheses equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis equals 0 rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals negative open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis rightwards double arrow open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus a b subscript 1 close parentheses left parenthesis b subscript 1 c minus a subscript 1 b right parenthesis

General

Maths-

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to

The two quadratic equations are

a x squared plus 2 c x plus b equals 0

and

a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis

Let

alpha

be the common root.
Now,

a x squared plus 2 c x plus b equals 0 rightwards double arrow a alpha squared plus 2 c alpha plus b equals 0 _ _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 1

a x squared plus 2 b x plus c equals 0 rightwards double arrow a alpha squared plus 2 b alpha plus c equals 0 _ _ _ _ _ _ _ _ e q u a t i o n 2

subtracting the equations, we get

2 alpha left parenthesis c minus b right parenthesis plus left parenthesis b minus c right parenthesis equals 0 rightwards double arrow 2 alpha equals fraction numerator c minus b over denominator c minus b end fraction rightwards double arrow alpha equals 1 half

now putting the value in equation1

a alpha squared plus 2 c alpha plus b equals 0 rightwards double arrow a open parentheses 1 half close parentheses squared plus 2 c 1 half plus b equals 0 rightwards double arrow a over 4 plus c plus b equals 0 rightwards double arrow a plus 4 b plus 4 c equals 0

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to

Maths-General

The two quadratic equations are

a x squared plus 2 c x plus b equals 0

and

a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis

Let

alpha

be the common root.
Now,

a x squared plus 2 c x plus b equals 0 rightwards double arrow a alpha squared plus 2 c alpha plus b equals 0 _ _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 1

a x squared plus 2 b x plus c equals 0 rightwards double arrow a alpha squared plus 2 b alpha plus c equals 0 _ _ _ _ _ _ _ _ e q u a t i o n 2

subtracting the equations, we get

2 alpha left parenthesis c minus b right parenthesis plus left parenthesis b minus c right parenthesis equals 0 rightwards double arrow 2 alpha equals fraction numerator c minus b over denominator c minus b end fraction rightwards double arrow alpha equals 1 half

now putting the value in equation1

a alpha squared plus 2 c alpha plus b equals 0 rightwards double arrow a open parentheses 1 half close parentheses squared plus 2 c 1 half plus b equals 0 rightwards double arrow a over 4 plus c plus b equals 0 rightwards double arrow a plus 4 b plus 4 c equals 0

General

physics-

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14 $m s to the power of negative 1 end exponent$ to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,

v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14 $m s to the power of negative 1 end exponent$ to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

physics-General

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,

v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

4C3 × 5C3 ×6C3 4P3× 5 P3 × 6 P3

Here, we can use the formula

The correct answer is: 4P3× 5 P3 × 6 P3

Related Questions to study

Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-

Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-

Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-

Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

A thin uniform annular disc (see figure) of mass has outer radius and inner radius . The work required to take a unit mass from point on its axis to infinity is

A thin uniform annular disc (see figure) of mass has outer radius and inner radius . The work required to take a unit mass from point on its axis to infinity is

The two bodies of mass and respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

The two bodies of mass and respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

If to terms, terms, , then

If to terms, terms, , then

The coefficient of in is

The coefficient of in is

Compounds (A) and (B) are –

Compounds (A) and (B) are –

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

If one root of the equation is reciprocal of the one of the roots of equation then

If one root of the equation is reciprocal of the one of the roots of equation then

If the quadratic equation and have a common root then is equal to

If the quadratic equation and have a common root then is equal to

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14 to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14 to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

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⁴C₃ × ⁵C₃ ×⁶C₃
⁴P₃×⁵P₃ × ⁶P₃
$\frac{15!}{4! 5! 6!}$
$\frac{15!}{(3!)^{3}}$

Here, we can use the formula
$C presuperscript n subscript r space cross times r factorial equals space fraction numerator n factorial over denominator left parenthesis n minus r factorial right parenthesis end fraction space space equals P presuperscript n subscript r$

The correct answer is: ⁴P₃×⁵P₃ × ⁶P₃

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

The coefficient of $x to the power of negative n end exponent$ in $left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent$ is

The coefficient of $x to the power of negative n end exponent$ in $left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent$ is

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

If one root of the equation $a x squared plus b x plus c equals 0$ is reciprocal of the one of the roots of equation $a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0$ then

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to

If the quadratic equation $a x squared plus 2 c x plus b equals 0$ and $a x squared plus 2 b x plus c equals 0 left parenthesis b not equal to c right parenthesis$ have a common root then $a plus 4 b plus 4 c$ is equal to