Maths-
General
Easy

Question

The number of ways in which 20 volunteers can be divided into groups of 4, 7 and 9 persons is-

  1. 16C7× 13C2    
  2. 20C7 × 11C4    
  3. 20C4 × 16C7    
  4. 20C9 × 13C9    

Hint:

Since  we have to make groups of 4, 7 , 9 people out of 20 volunteers , find out the number of people that can be selected and similarly those who are selected in one group , they will not participate in other groups.

The correct answer is: 20C4 × 16C7


    We have to make groups of 4, 7 , 9 people out of 20 volunteers
    4 people out of 20 can be selected in C presuperscript 20 subscript 4 space w a y s.
    7 people out of remaining 16 can be selected in C presuperscript 16 subscript 7 space w a y s.
    9 people out of remaining 9 can be selected in C presuperscript 9 subscript 9 space w a y s space equals space 1.

    Thus, the number of ways in which 20 volunteers can be divided into groups of 4, 7 and 9 persons is
    C presuperscript 20 subscript 4 space cross times C presuperscript 16 subscript 7 space.

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    Detailed Solution :
    5 heaps of 3 books each are to be made from 15 different books. We are to find in how many ways this can be done.
    We know that the formula of dividing different things into groups of sizes a subscript 1 comma space a subscript 2 comma a subscript 3 space end subscript........ a subscript n space where
    a subscript 1 plus space a space plus space a subscript 3 space space end subscript plus space........ plus a subscript n space = fraction numerator m factorial over denominator left parenthesis a subscript 1 factorial right parenthesis left parenthesis a subscript 2 factorial right parenthesis left parenthesis a subscript 3 factorial right parenthesis......... left parenthesis a subscript n factorial right parenthesis end fraction
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    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

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    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

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    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
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    parallel

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