Maths-
General
Easy

Question

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

  1. 10P4    
  2. 10C4    
  3. 4    
  4. 40    

Hint:

Use formula
space C presuperscript n subscript r space cross times space r factorial space equals space space P presuperscript n subscript r

The correct answer is: 10P4


    You are given 10 flags of different colors and you can give signals, but use only 4 flags.


    First you have to choose 4 flags from 10 flags
    This can be done in C presuperscript 10 subscript 4 space end subscript ways
    Among this 4 flags of different colors, 4! ways of different arrangements are possible
    So total number of signals = C presuperscript 10 subscript 4 x space 4 factorial
    S i n c e space C presuperscript n subscript r space cross times space r factorial space equals space space P presuperscript n subscript r
    
C presuperscript 10 subscript 4 x space 4 factorial equals equals P presuperscript 10 subscript 4

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    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-

    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-

    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-

    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!×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×(8× 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
     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!×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×(8× 8!)
    General
    Maths-

    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-

    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

    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

    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

    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 – 

    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 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
    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:

    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

    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
    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

    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