Maths-

General

Easy

Question

The number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is -

⁵C₃× ⁴C₂ × 3! × 2!
⁵P₃ × ⁴P₂ × 5!
⁵C₃ × ⁴C₂ × 5!
⁵C₃ × ⁴C₂ × $\frac{5!}{2!}$

Hint:

No repetition is allowed.
Number of ways of taking any 2 digits from digits 6,7,8,9 = $C presuperscript 4 subscript 2$ similarly find out the rest.

The correct answer is: ⁵C₃ × ⁴C₂ × 5!

We have to form a 5digit number by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5
Number of ways of taking any 2 digits from digits 6,7,8,9 = $C presuperscript 4 subscript 2$
Number of ways of taking any 3 digits from 1, 2, 3, 4, 5 = $C presuperscript 5 subscript 3$
Number of ways of forming a 5 digit number = 5!
Thus, the number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is - $C presuperscript 4 subscript 2$ $cross times$ $C presuperscript 5 subscript 3$ $cross times$ 5! .

How many numbers consisting of 5 digits can be formed in which the digits 3,4 and 7 are used only once and the digit 5 is used twice-

Maths-General

General

Maths-

The number of ways of distributing n prizes among n boys when any of the student does not get all the prizes is-

Maths-General

General

Maths-

If the line $x plus y plus k equals 0$ is a tangent to the parabola $x to the power of 2 end exponent equals 4 y$ then k=

Maths-General

General

Maths-

The number of ways in which n prizes can be distributed among n students when each student is eligible to get any number of prizes is-

Maths-General

General

Maths-

In how many ways can six different rings be wear in four fingers?

Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.

Now, we know that by the fundamental principle of counting there can be

4 \times 4 \times 4 \times 4 \times 4 \times 4

ways of wearing 6 rings.
So, we have

4 to the power of 6

4^{6} = 4096

ways to wear 6 different types of rings.

In how many ways can six different rings be wear in four fingers?

Maths-General

4 \times 4 \times 4 \times 4 \times 4 \times 4

ways of wearing 6 rings.
So, we have

4 to the power of 6

4^{6} = 4096

ways to wear 6 different types of rings.

General

Maths-

How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

The number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is -

5C3 × 4C2 × 3! × 2! 5P3 × 4P2 × 5! 5C3 × 4C2 × 5! 5C3 × 4C2 ×

No repetition is allowed.Number of ways of taking any 2 digits from digits 6,7,8,9 = similarly find out the rest.

The correct answer is: 5C3 × 4C2 × 5!

Related Questions to study

How many numbers consisting of 5 digits can be formed in which the digits 3,4 and 7 are used only once and the digit 5 is used twice-

How many numbers consisting of 5 digits can be formed in which the digits 3,4 and 7 are used only once and the digit 5 is used twice-

The number of ways of distributing n prizes among n boys when any of the student does not get all the prizes is-

The number of ways of distributing n prizes among n boys when any of the student does not get all the prizes is-

If the line is a tangent to the parabola then k=

If the line is a tangent to the parabola then k=

The number of ways in which n prizes can be distributed among n students when each student is eligible to get any number of prizes is-

The number of ways in which n prizes can be distributed among n students when each student is eligible to get any number of prizes is-

In how many ways can six different rings be wear in four fingers?

In how many ways can six different rings be wear in four fingers?

How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

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

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

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

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⁵C₃× ⁴C₂ × 3! × 2!
⁵P₃ × ⁴P₂ × 5!
⁵C₃ × ⁴C₂ × 5!
⁵C₃ × ⁴C₂ × $\frac{5!}{2!}$

No repetition is allowed.
Number of ways of taking any 2 digits from digits 6,7,8,9 = $C presuperscript 4 subscript 2$ similarly find out the rest.

The correct answer is: ⁵C₃ × ⁴C₂ × 5!

If the line $x plus y plus k equals 0$ is a tangent to the parabola $x to the power of 2 end exponent equals 4 y$ then k=

If the line $x plus y plus k equals 0$ is a tangent to the parabola $x to the power of 2 end exponent equals 4 y$ then k=

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