What is the flaw in my logic for this question? Q. Find the number of different committees of $5$ that can be chosen from $7$ boys and $5$ girls, if they include at least one of each sex.
My logic was $(7C1)$ [choose at least 1 boy] times
$(5C1)$ [choose at least one girl] times $(10C3)$ [the rest]/
I know the correct was could be $(7C4)(5C1) + (7C3)(5C2) + \ldots +(7C1)(5C4)$.  However, I can't see the flaw in my logic. This is probably because I don't have a strong understanding of combinatorial thinking yet.
 A: Imagine John being part of the committee. You might have selected him out of the seven men at first, and then have put three other people with him afterward (let's call them Adam, Mark and Sarah). Alternatively, it is possible that you selected Adam first, but put John in as part of your second selection process (which also included Mark and Sarah). Because both committees are the same, you have indeed overcounted. Note that you are not overcounting in the alternative approach, because each person is selected (or not) in one single step.
A: You first choose $1$ from the boys and $1$ from the girls. You then choose $3$ from rest$10$.
Let's take cases of overcounting of boys and look at overcounting of girls at the same time -
Case1 - one boy in the committee - $7$ ways. There is no overcounting of boys but you have $4$ girls in the committee overcounted by $15$ and multiplied by $7$ ways of choosing boys.
Overcounting = 105
Case2 - All committees where you have $2$ boys and rest of them girls, you have already chosen all the boys as the first boy with $^7C_1$. Now you choose the second boy from rest $6$. So you will get $42$ ways in total. But the way to choose two boys is $^7C_2 = 21$. You are counting each combination twice. Then you multiply this by number of ways to choose $3$ girls to the committee so it multiplies and which btw has been overcounted too (by 20).
Overcounting = 3042 - 1021 = 1050
Case3 - $3$ boys. You count it as $^7C_1\times ^6C_2 = 105$ whereas there are only $^7C_3 = 35$ ways. So overcounted by $70$. Again the multiplication effect and double counting of choosing $2$ girls.
Overcounting = 1750
Case4 - $4$ boys. You count it as $^7C_1\times ^6C_3 = 140$ whereas there are only $^7C_4 = 35$ ways. So overcounted by $105$ ways. No overcounting of girls as there is only one but overcounting of boys multiplied by $5$ as there are $5$ ways to choose a girl.
Overcounting = 525
Every such case will have overcounting and then the multiplication effect of these overcounting.
Total overcounting = 3430
Number of ways you counted = $4200$
Number of ways that should be counted = $770$
Difference = $3430$.
A: Options with multiple boys and/or multiple girls can be chosen more than once, and therefore your method is guilty of overcounting.
A: I'll use an example to show how the overcounting happens.
Let's name the boys: A, B, C, D, E, F, G
And the girls: H, I, J, K, L
In your logic you first pull out a boy and girl and then count the rest of the combinations. So your logic would count AH/BCD and BH/ACD as two separate combinations even though they are the same group of 5.
