7 boys and 4 girls, to form a committee of 6 when the committee needs exactly 2 girls. Find the number of ways.

Question

This question is very easy indeed as we just take $${4 \choose 2} \times {7 \choose 4} = 210$$

However, let us assume for a moment that we are also interested in the possible arrangements of the members of the committee: That is we care about ordering. Now being someone who is super bad at combinatorics: My question is: Should we choose $${4 \choose 2} \times {7 \choose 4} \times 6!$$ or $${4 \choose 2}\times 2! \times {7 \choose 4} \times 4!$$

I think we should choose the former (first one). However i fail to understand why the second one doesn't work.

Answer 1

There are $7$ boys and $4$ girls.

Then we must choose $2$ girls, and then choose $4$ boys, and arrange the $6$ of them.

Hence we have

\begin{align} {4 \choose 2} \cdot {7 \choose 4} \cdot 6! \end{align}

The latter does not work because you would be separately arranging the girls and then arranging boys, which does not amount to anything significant. We want to arrange the six of them together once we choose the committee. That's why we must multiply by $6!$.

Answer 2

Hint

In the second one we need to select the places we want the boys or girls to appear too. Hence the answer can be given as $$\binom {4}{2}\cdot 2!\cdot \binom {7}{4}\cdot \binom {6}{4}\cdot 4!$$

We multiply the extra term $\binom {6}{4}$ to choose the place at which the boy will be selected( e.g. The boys can be selected on 1,3,4,5 positions while selecting the committee members, etc) . Since we select 4 places for boys, 2 places for girls get automatically selected.

Answer 3

yeah, you're right former case is correct and Latter is wrong

in former case after you've selected 4 boys and 2 girls and you permute all of them togeather in $6!$ ways

whereas in latter case permutation occurs only within 4 boys group$(B_{1}B_{2}B_{3}B_{4})$ with no girl between them,

and within 2 girls group$(G_{1} G_{2})$ with no boys between them

so here you're missing ordering like $B_{1}G_{2}B_{2}B_{4}G_{1}B_{3}$ so latter is wrong and former is correct.