need help setting up a probability question

Question

A committee of five is to be selected from a group of 12 children. The two youngest cannot both be on the committee. In how many ways can the committee be selected?

I thought this would be ${^{10}\mathrm C_5}\cdot{^2\mathrm C_1}$, but that is not the case. How would I set this up?

Answer 1

$\def\cbinom#1#2{{^{#1}\mathrm C_{#2}}}\cbinom{10}4\cdot\cbinom 21$ counts ways to set up the committee such that one from the two (and four from the other ten) are on it .   That is okay but it is not all you want to count.

You want to count ways to set up the committee of five such that one or none from the two are on it (that is: not both from the two) and the remainder selected from the other ten.

So you must add the count of ways to select none from the two and five from the remaining ten.

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Alternatively: count the total ways to select five from twelve, and subtract the ways to select three from ten and both from the two.

Answer 2

There are $\binom{12}{5}$ total combinations. There are $\binom{10}{3}$ combinations including both youngest. Final result $\binom{12}{5}-\binom{10}{3}$.