Permutations And Combinations

Question
CBSEENMA11014202

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done, when the committee consists of :

(i) exactly 3 girls? (ii) at least three girls? (iii) at most 3 girls?

Solution

Number of boys = 9
Number of girls = 4
The committee is to consist of 7
(i) The committee consists of exactly 3 girls
rightwards double arrow                 Number of boys = 7 - 3 = 4
∴              The number of selections = straight C presuperscript 9 subscript 4 cross times straight C presuperscript 4 subscript 3 space equals space fraction numerator 9 factorial over denominator 4 factorial space 5 factorial end fraction space cross times space fraction numerator 4 factorial over denominator 3 factorial space 1 factorial end fraction
                                                 
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(ii) The committee consists of at least 3 girls.
rightwards double arrow  The committee consists of 3 girls + 4 boys or 4 girls + 3 boys
                                 (Since, the number of girl is 4, so there is no further option)
When committee consists of 3 girls + 4 boys
                      Number of selections = straight C presuperscript 4 subscript 3 cross times straight C presuperscript 9 subscript 4 space equals space 504
When committee consists of 4 girls + 3 boys
                      Number of selections = straight C presuperscript 4 subscript 4 cross times straight C presuperscript 9 subscript 3 space equals space 1 cross times fraction numerator 9 cross times 8 cross times 7 cross times 6 factorial over denominator 6 factorial space space 3 factorial end fraction space equals space 84.
Hence, the number of selections for the committee = 504 + 84 = 588.
(iii) The committee consists of at most 3 girls.
      This gives us the following options:
                       No girl + 7 boys
                     
                  Number of selections = space space straight C presuperscript 4 subscript 0 cross times straight C presuperscript 9 subscript 7 space equals space 1 cross times fraction numerator 9 factorial over denominator 2 factorial space space 7 factorial end fraction space equals space 36.
                                 Or
                       1 girl + 6 boys
                 Number of selections = straight C presuperscript 4 subscript 1 cross times straight C presuperscript 9 subscript 6 space equals space 4 cross times fraction numerator 9 cross times 8 cross times 7 cross times 6 factorial over denominator 6 factorial space space 3 factorial end fraction space equals space 4 cross times 84 equals 336.
                                 Or
                            2 girls + 5 boys
                   Number of selections = straight C presuperscript 4 subscript 2 cross times straight C presuperscript 9 subscript 5 space equals space fraction numerator 4 factorial over denominator 2 factorial space 2 factorial end fraction cross times fraction numerator 9 factorial over denominator 4 factorial space 5 factorial end fraction
                                               
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                                Or
                            3 girls + 4 boys
                    Number of selections = space space straight C presuperscript 4 subscript 3 cross times straight C presuperscript 9 subscript 4 space equals space 4 cross times fraction numerator 9 cross times 8 cross times 7 cross times 6 cross times 5 factorial over denominator 5 factorial space 4 factorial end fraction equals 504.
Hence, the total number of selections = 36 + 336 + 756 + 504 = 1632.
              
 
 

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