TITLE: Combined Combinations - What's wrong with my line of thinking?
QUESTION [5 upvotes]: I'm brushing up on some basics for an upcoming job interview, and I've got myself confused over a question about combined combinations.
From Discrete Mathematics with Graph Theory Third Edition by Goodaire & Parmenter, Section 7.2, Exercise 1:

A group of people is comprised of six from Nebraska, seven from Idaho,
and eight from Louisiana.
(a) In how many ways can a committee of six be formed with two people from each state?
(b) In how many ways can a committee of seven be formed with at least two people from each state?

I'm fine with part (a). I got the same solution as is in the back of the book:
$\binom{6}{2} \binom{7}{2} \binom{8}{2} = 8820$
But with part (b), I was thinking that the answer would be the result from part (a), multiplied by the number of ways you could choose one more person from the total pool of remaining people:
$\binom{6}{2} \binom{7}{2} \binom{8}{2} \binom{15}{1} = 132,300$
The answer in the back of the book is
$\binom{6}{3} \binom{7}{2} \binom{8}{2} + \binom{6}{2} \binom{7}{3} \binom{8}{2} + \binom{6}{2} \binom{7}{2} \binom{8}{3} = 44,100$
I haven't touched this stuff in a while, so I'm inclined to believe the authors, but my logic still makes sense to me.
What is the problem with the way I'm thinking about the answer?

REPLY [2 votes]: Try the same problem with smaller numbers! (This can help you with a lot of combinatorial problems.)
A group of people is composed of two from Nebraska and two from Idaho.
(a) In how many ways can a committee of two be formed with one person from each state?
$$\binom21\binom21=4$$
(b) In how many ways can a committee of three be formed with at least one person from each state?
In this case, any committee of three will include at least one person from each state, so the correct answer is
$$\binom43=\binom41=4$$
On the other hand, by your line of thinking, the answer would be the result from part (a), multiplied by the number of ways you could choose one more person from the remaining pool of people:
$$4\cdot\binom21=8$$
Clearly, the second answer is wrong; we have overcounted somehow. Because the numbers are so small, you can actually write down the $8$ committees you have counted, and see for yourself how the overcount happened. Call the Nebraskans $N,E$ and the Idahoans $I,D.$
$$NI\longrightarrow NIE,\ NID$$
$$ND\longrightarrow NDE,\ NDI$$
$$EI\longrightarrow EIN,\ EID$$
$$ED\longrightarrow EDN,\ EDI$$
Observe that, since order doesn't matter, $NID$ and $NDI$ are the same committee, which has been counted twice.