Question

A committee consisting of at least three members is to be formed from a group of 6 boys and 6 girls such that it always has a boy and girl. The number of ways to form such a committee is___________.

Answer 1

Hint: The number of $k$-combinations for all $k$ is the number of subsets of a set of $n$ elements. There are several ways to see that this number is $2^n$. We use this and find the values of probabilities of having no boys and girls and finally subtract it from the probabilities of having a boy and girl and then find the required combinations.

Complete step by step solution:
It is given that the committee consisting of at least three members is to be formed from a group of 6 boys and 6 girls such that it always has a boy and girl.
The total number of committees can be formed from 6 boys and 6 girls i.e. from 12 members is \[{2^{12}}\] if the members are chosen randomly without the restriction of sex or the number of members.
The total number of committees that can be formed from 6 boys is \[{2^6}\] (no girls are chosen for this case)
The total number of committees that can be formed from 6 girls is \[{2^6}\] (no boys are chosen for this case)
So the total number of committees that can be formed from 6 boys and that can be formed from 6 girls is = \[{2^6}\]+\[{2^6}\]=\[{2.2^6} = {2^7}\]
But the committee that no girls are selected or no boys are selected is taken twice, so we have to delete one from that.
So required number of committee =\[{2^7} - 1\]
Now, if we exclude these from the total number of committees we can exclude the committees consist one member, so the number of such committees is =\[{2^{12}}\]-(\[{2^7} - 1\])
Now we have to find out the number of committees for which at least three members are present.
So the number of ways to choose 1 boy from 6 boys is \[^6{C_1}\] and the number of ways to choose 1 girl from 6 girls is \[^6{C_1}\].
So number of ways to choose 1 boy and one girl member is =\[^6{C_1}\]×\[^6{C_1}\]
Here we are going to use the formula of combinations to find the number of ways, the formula for finding combinations is given by, \[C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
=\[\dfrac{{6!}}{{5!1!}} \times \dfrac{{6!}}{{5!1!}} = 6 \times 6 = 36\]
The committee consisting of at least three members such that it always has a boy and girl =
\[ = {2^{12}} - ({2^7} - 1) - 36\]
\[ = {2^{12}} - {2^7} + 1 - 36\]
By expressing the terms in power we get,
\[ = 4096 - 128 - 35\]\[ = 3933\]

$\therefore$ The number of ways to form the committee is \[3933\].

Note:
Here we are using the fact that initially, we are going to find the number of committees with no boy and girl and subtract it from the total number of the committee so that it would become much easy to find a number of ways.