Number of ways of choosing seven children from a classroom of 32 (15 boys, 17 girls) with at least 1 boy

I know that the correct solution can be calculated as: $$ \binom {32} {7} - \binom {17}{7}$$

But why is the following solution incorrect? (I am interested in why the following reasoning is incorrect, I realize that the two numbers are not equal): $$ \frac{15 \binom {31} {6}}{2!} $$

The reasoning is that we first pick a boy ($15$ options) and then pick $6$ children out of $31$ remaining in an arbitrary manner. Finally divide by $2!$ since the order of the two groups does not matter.

Solution 1:

Let's enumerate boys as $B_1,B_2,...,B_{15}$ and girls as $G_1,G_2,...,G_{17}$. Then in your method, we choose one of them first, say $B_1$. Then for the remaining $6$ children, say $B_2$ is among them and the rest is $G_1,G_2,...,G_5$. But this is as same as choosing $B_2$ first and then choosing other $6$ children as $B_1,G_1,G_2,...,G_5$. Although this is only one example, you can easily see that in your method we are overcounting.

Solution 2:

Division by 2 is uncorrect, as an alternative you could calculate it as

$$\binom {15}{1}\binom {17}{6}+\binom {15}{2}\binom {17}{5}+\binom {15}{3}\binom {17}{4}+\binom {15}{4}\binom {17}{3}+\binom {15}{5}\binom {17}{2}+\binom {15}{6}\binom {17}{1}+\binom {15}{7}\binom {17}{0}=\sum_{k=1}^7\binom{15}{k}\binom{17}{7-k}$$

showing also that by counting principle

$$\sum_{k=1}^7\binom{15}{k}\binom{17}{7-k}=\binom{32}{7} - \binom{17}{7}$$

and more in general the Vandermonde's identity

$$\binom{m+n}{r}=\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}$$