A committee of 7 to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of
A) Exactly 3 girls?
B) At least 3 girls?
C) At most 3 girls?
Answer
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Hint: This is a permutation combination question. For the first case there will be only 1 type of possibility i.e. of 3 girls and 4 boys. For second case 2 possibility will arise i.e. having 3 girls and 4 boys, 4 girls and 3 boys respectively. Similarly for the third case there will be 4 possibilities i.e. of having girls number 3, 2, 1 and 0 respectively for each case. And using combination rules we will derive the exact number of ways of selection for each case.
Complete step-by-step answer:
According to the question,
A committee to be formed of 7 members.
Number of boys = 9
Number of girls =4
Case (i)
The committee consists of exactly 3 girls.
Hence, the number of boys will be in the committee $ = 7 - 3 = 4 $
$ \therefore $ The number of ways the selection can be done of 3 girls and 4 boys = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
$ \therefore $ In 504 ways selection can be done when the committee consists of exactly 3 girls.
Case (ii)
The committee is to consist of at least 3 girls that means it can have at most 4 girls as there are a total 4 girls.
For that two types of cases arise. I.e. committee consists of 3 girls and $ 7 - 3 = 4 $ boys
And the second one is committee consists of 4 girls and $ 7 - 4 = 3 $ boys.
$ \therefore $ Number of ways 3 girls and 4 boys to be selected = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
And number of ways 4 girls and 3 boys to be selected = $ {}^9{C_3} \times {}^4{C_4} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_3} \times {}^4{C_4} $
$ = \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}} \times \dfrac{{4!}}{{4!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{3!6!}} \times 1 $
Writing 9! In terms of 6! we get,
$ = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3!6!}} $
Cancelling 6! From numerator and denominator,
$ = \dfrac{{504}}{6} = 84 $
$ \therefore $ The total number of way selection can be done when committee consists of at least 3 girls is $ = 504 + 84 = 588 $
Hence, In 588 ways selection can be done when the committee consists of at least 3 girls.
Case (iii)
The committee is to be consists of at most 3 girls that means it can have at least 0 girls
For that 4 types of cases arise. I.e. committee consists of 3 girls and $ 7 - 3 = 4 $ boys
Second one is committee consists of 2 girls and $ 7 - 2 = 5 $ boys
Third one is committee consists of 1 girl and $ 7 - 1 = 6 $ boys.
And the fourth one is committee consists of 0 girls and 7 boys.
$ \therefore $ Number of ways 3 girls and 4 boys to be selected = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
$ \therefore $ Number of ways 2 girls and 5 boys to be selected = $ {}^9{C_5} \times {}^4{C_2} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_5} \times {}^4{C_2} $
$ = \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{5!4!}} \times \dfrac{{4!}}{{2!2!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{2 \times 1 \times 2 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{4} = 756 $
$ \therefore $ Number of ways 1 girl and 6 boys to be selected = $ {}^9{C_6} \times {}^4{C_1} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_6} \times {}^4{C_1} $
$ = \dfrac{{9!}}{{6!\left( {9 - 6} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{6!3!}} \times \dfrac{{4!}}{{1!3!}} $
Writing 9! In terms of 6! And 4! In terms of 3! we get,
$ = \dfrac{{9 \times 8 \times 7 \times 6!}}{{6!3!}} \times \dfrac{{4 \times 3!}}{{1 \times 3 \times 2 \times 1}} $
Cancelling 6! and 3! from numerator and denominator,
$ = \dfrac{{9 \times 8 \times 7 \times 4}}{6} = \dfrac{{2016}}{6} = 336 $
$ \therefore $ Number of ways 0 girl and 7 boys (only 7 boys) to be selected = $ {}^9{C_7} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_7} $
$ = \dfrac{{9!}}{{7!\left( {9 - 7} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{7!2!}} $
Writing 9! In terms of 7! we get,
$ = \dfrac{{9 \times 8 \times 7!}}{{7! \times 2}} $
Cancelling 7! from numerator and denominator,
$ = \dfrac{{72}}{2} = 36 $
$ \therefore $ The total number of way selection can be done when committee consists of at most 3 girls is $ = 504 + 756 + 336 + 36 = 1632 $
Hence, In 1632 ways selection can be done when the committee consists of at most 3 girls.
Note: A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Its formula is given by, $ P(n,r) = n!(n - r)! $
A combination is a technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter. Its formula is given by, $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ .
When the order does not matter, it is a combination.
When order does matter, it is a permutation.
Complete step-by-step answer:
According to the question,
A committee to be formed of 7 members.
Number of boys = 9
Number of girls =4
Case (i)
The committee consists of exactly 3 girls.
Hence, the number of boys will be in the committee $ = 7 - 3 = 4 $
$ \therefore $ The number of ways the selection can be done of 3 girls and 4 boys = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
$ \therefore $ In 504 ways selection can be done when the committee consists of exactly 3 girls.
Case (ii)
The committee is to consist of at least 3 girls that means it can have at most 4 girls as there are a total 4 girls.
For that two types of cases arise. I.e. committee consists of 3 girls and $ 7 - 3 = 4 $ boys
And the second one is committee consists of 4 girls and $ 7 - 4 = 3 $ boys.
$ \therefore $ Number of ways 3 girls and 4 boys to be selected = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
And number of ways 4 girls and 3 boys to be selected = $ {}^9{C_3} \times {}^4{C_4} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_3} \times {}^4{C_4} $
$ = \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}} \times \dfrac{{4!}}{{4!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{3!6!}} \times 1 $
Writing 9! In terms of 6! we get,
$ = \dfrac{{9 \times 8 \times 7 \times 6!}}{{3!6!}} $
Cancelling 6! From numerator and denominator,
$ = \dfrac{{504}}{6} = 84 $
$ \therefore $ The total number of way selection can be done when committee consists of at least 3 girls is $ = 504 + 84 = 588 $
Hence, In 588 ways selection can be done when the committee consists of at least 3 girls.
Case (iii)
The committee is to be consists of at most 3 girls that means it can have at least 0 girls
For that 4 types of cases arise. I.e. committee consists of 3 girls and $ 7 - 3 = 4 $ boys
Second one is committee consists of 2 girls and $ 7 - 2 = 5 $ boys
Third one is committee consists of 1 girl and $ 7 - 1 = 6 $ boys.
And the fourth one is committee consists of 0 girls and 7 boys.
$ \therefore $ Number of ways 3 girls and 4 boys to be selected = $ {}^9{C_4} \times {}^4{C_3} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_4} \times {}^4{C_3} $
$ = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{4!5!}} \times \dfrac{{4!}}{{3!1!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{3 \times 2 \times 1 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{6} = 504 $
$ \therefore $ Number of ways 2 girls and 5 boys to be selected = $ {}^9{C_5} \times {}^4{C_2} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_5} \times {}^4{C_2} $
$ = \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{5!4!}} \times \dfrac{{4!}}{{2!2!}} $
Cancelling 4! From numerator and denominator and writing 9! In terms of 5! We get,
$ = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} \times \dfrac{1}{{2 \times 1 \times 2 \times 1}} $
Cancelling 5! From numerator and denominator,
$ = \dfrac{{3024}}{4} = 756 $
$ \therefore $ Number of ways 1 girl and 6 boys to be selected = $ {}^9{C_6} \times {}^4{C_1} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_6} \times {}^4{C_1} $
$ = \dfrac{{9!}}{{6!\left( {9 - 6} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{6!3!}} \times \dfrac{{4!}}{{1!3!}} $
Writing 9! In terms of 6! And 4! In terms of 3! we get,
$ = \dfrac{{9 \times 8 \times 7 \times 6!}}{{6!3!}} \times \dfrac{{4 \times 3!}}{{1 \times 3 \times 2 \times 1}} $
Cancelling 6! and 3! from numerator and denominator,
$ = \dfrac{{9 \times 8 \times 7 \times 4}}{6} = \dfrac{{2016}}{6} = 336 $
$ \therefore $ Number of ways 0 girl and 7 boys (only 7 boys) to be selected = $ {}^9{C_7} $
Expanding the right hand side of the above equation as per the combination formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we get,
R.H.S- $ {}^9{C_7} $
$ = \dfrac{{9!}}{{7!\left( {9 - 7} \right)!}} $
Again simplifying it we get,
$ = \dfrac{{9!}}{{7!2!}} $
Writing 9! In terms of 7! we get,
$ = \dfrac{{9 \times 8 \times 7!}}{{7! \times 2}} $
Cancelling 7! from numerator and denominator,
$ = \dfrac{{72}}{2} = 36 $
$ \therefore $ The total number of way selection can be done when committee consists of at most 3 girls is $ = 504 + 756 + 336 + 36 = 1632 $
Hence, In 1632 ways selection can be done when the committee consists of at most 3 girls.
Note: A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Its formula is given by, $ P(n,r) = n!(n - r)! $
A combination is a technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter. Its formula is given by, $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ .
When the order does not matter, it is a combination.
When order does matter, it is a permutation.
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