Question

In how many ways we can select a committee of $6$ persons from $6$ boys and $3$ girls. If at least two boys and at least two girls must be in the committee?

Answer 1

Hint: Combination is the mathematical technique to find the total number of ways formed to arrange the given set of conditions. In this, we find the total number of arrangements where order doesn't matter. For example: if we choose r out of n items, then the total number of ways are \[{}^n{C_r} = \dfrac{{n!}}{{(n - r)! \cdot r!}}\]
Factorial: \[n! = n(n - 1)(n - 2).......3.2.1\]
Factorial is denoted as !
Complete step-by-step answer:
We have to make a committee of $6$ persons which are selected from $6$ boys and $3$ girls and there is a condition that in the committee, at least $2$ girls and $2$ boys should be there.
Then, how many ways are possible in which these 6 persons can be selected.
Therefore, the total number of ways is equal to the sum of following $2$ factors.
At least $2$ girls will be there, it means total $6$$ = 2$ girls $4$ boys
Also, if $3$ girls will be there, it means total $6 = 3$ girls $3$ boys
Because in at least $2$ girls, the condition of $2$ girls and $3$ girls is there.
$\therefore $Total number of ways $ = $ $2$ girls $4$ boys $ + $ $3$ girls $3$ boys
$ = {}^3{C_2} \times {}^6{C_4} + {}^3{C_3} \times {}^6{C_3}$
Using formula of Combination, we will solve these
\[{}^n{C_r} = \dfrac{{n!}}{{(n - r)! \cdot r!}}\]
\[ = {}^3{C_2} \times {}^6{C_4} + {}^3{C_3} \times {}^6{C_3}\]
\[ = \dfrac{{3!}}{{2!1!}} \times \dfrac{{6!}}{{4!2!}} + \dfrac{{3!}}{{1!3!}} \times \dfrac{{6!}}{{3!3!}}\]
$ = 3 \times 15 + 1 \times 20$
$ = 45 + 20$
$ = 65$

$\therefore $There are $65$ ways to select girls and boys to form a $6$ members committee.

Note: Factorial of $0$ is $1$, i.e. 0!=1. Factorials of negative numbers cannot be calculated. Along with the combination , permutation is also a technique in which we calculate the total number of arrangements but here order matters and formula of permutation for choosing ‘$r$’ out of ‘$n$’ is \[{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}\].