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Question: In how many ways we can select a committee of \(6\) persons from \(6\) boys and \(3\) girls. If at l...

In how many ways we can select a committee of 66 persons from 66 boys and 33 girls. If at least two boys and at least two girls must be in the committee?

Explanation

Solution

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 nCr=n!(nr)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)! \cdot r!}}
Factorial: n!=n(n1)(n2).......3.2.1n! = n(n - 1)(n - 2).......3.2.1
Factorial is denoted as !
Complete step-by-step answer:
We have to make a committee of 66 persons which are selected from 66 boys and 33 girls and there is a condition that in the committee, at least 22 girls and 22 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 22 factors.
At least 22 girls will be there, it means total 66 =2 = 2 girls 44 boys
Also, if 33 girls will be there, it means total 6=36 = 3 girls 33 boys
Because in at least 22 girls, the condition of 22 girls and 33 girls is there.
\therefore Total number of ways == 22 girls 44 boys ++ 33 girls 33 boys
=3C2×6C4+3C3×6C3= {}^3{C_2} \times {}^6{C_4} + {}^3{C_3} \times {}^6{C_3}
Using formula of Combination, we will solve these
nCr=n!(nr)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)! \cdot r!}}
=3C2×6C4+3C3×6C3= {}^3{C_2} \times {}^6{C_4} + {}^3{C_3} \times {}^6{C_3}
=3!2!1!×6!4!2!+3!1!3!×6!3!3!= \dfrac{{3!}}{{2!1!}} \times \dfrac{{6!}}{{4!2!}} + \dfrac{{3!}}{{1!3!}} \times \dfrac{{6!}}{{3!3!}}
=3×15+1×20= 3 \times 15 + 1 \times 20
=45+20= 45 + 20
=65= 65

\therefore There are 6565 ways to select girls and boys to form a 66 members committee.

Note: Factorial of 00 is 11, 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 ‘rr’ out of ‘nn’ is nPr=n!(nr)!{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}.