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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?

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    $\begingroup$ You are over counting. Choosing $A,B$ from Nebraska and then choosing a third $C$ from there is the same as choosing $A,C$ and then choosing $B$ as the extra person. Indeed you are over counting by a factor of exactly $3$. $\endgroup$ – lulu Oct 24 '15 at 18:53
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You are triple-counting.

The person chosen in the $\binom{15}{1}$ could be Carol from Nebraska, and the people counted in the $\binom{6}{2}$ could be Alicia and Beti from Nebraska. This is the same group from Nebraska as the group obtained by choosing Alicia from Nebraska in the $\binom{15}{1}$, and the other two in the $\binom{6}{2}$. Same with Beti as the "third" person from Nebraska.

The point is that if we are choosing $3$ people from Nebraska, it is a bag of $3$, it is not choosing a pair and then choosing another.

Remark: In this case, we can deliberately triple count, and obtain your number. However, if we choose to do that, we must at the end divide by $3$. The strategy of controlled multiple-counting is often useful.

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  • $\begingroup$ D'oh. Of course, divide by three to avoid the triple count. I got bogged down in trying to cut down the the triple count that I just restarted from scratch to do it the book's way. $\endgroup$ – fleablood Oct 24 '15 at 19:03
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    $\begingroup$ "it is a bag of 3, it is not choosing a pair and then choosing another" -- this line really drove it home for me. Thanks. $\endgroup$ – WhiteHotLoveTiger Oct 24 '15 at 19:15
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    $\begingroup$ You are welcome. This sort of multiple counting will happen to you a few times, and then probably never again. A canonical example of double-counting comes when we are counting "two pair" poker hands. It is tempting to say that the kind of the "first" pair can be chosen in $\binom{13}{1}$ ways, and then the kind of the "second" pair can be chosen in $\binom{12}{1}$ ways. But there is no "first" pair and "second" pair. there are $\binom{13}{2}$ ways to choose the two kinds. Or do it the wrong way and then divide by $2$. $\endgroup$ – André Nicolas Oct 24 '15 at 19:24
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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.

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My first thought was to think about it exactly as you did. But the problem is we double count our results.

If you choose Bob and Jane from Nebraska, and then as our seventh member we randomly choose Jim also from Nebraska, that is the same result as choosing Bob and Jim from Nebraska and randomly choosing Jane as the seventh member. We counted those results twice.

Now, I can go back and try to figure out how many times we'd repeat these and adjust. [edit: As Jane, Jim + Bob = Jane,Bob + Jim = Bob, Jim + Jane, the adjusting is to divide by 3. I totally spaced on how to do this when I first posted this answer.] Or I could think: How many ways to have three from Nebraska and two from the other two states [$ {6 \choose 3} {7 \choose 2} {8 \choose 2}$], plus how many ways to have three from Idaho and two from the other[$ {6 \choose 2} {7 \choose 3} {8 \choose 2}$], plus how many ways to have three from Louisiana and two from the other two states[$ {6 \choose 2} {7 \choose 2} {8 \choose 3}$].

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  • $\begingroup$ Thanks. Listing out all three ways you can end up with Jane, Jim & Bob was helpful. $\endgroup$ – WhiteHotLoveTiger Oct 24 '15 at 19:20

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