Choosing a committee (combination problem) A group of people is comprised of six from Nebraska, seven from Idaho, and eight from Louisiana.


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*In how many ways can a committee of six be formed with two people from each state?


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*${^6C_2} \times {^7C_2} \times {^8C_2}\ $ Is this correct  ??


*In how many ways can a committee of seven be formed with at least two people in each state?


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*$\left({^6C_2}\right)(4!) \times \left({^7C_2}\right)(5!) \times \left({^8C_2}\right)(6!)$ 



I am actually not sure how to do this one. My intuition on this was that there must be at least $2$ from each state and I don't believe the order matters. Therefore it would be ${^6C_2} \times {^7C_2} \times {^8C_2}\ $ and then since there needs to be $7$ in total, there is a $4!$ possibility for the people from Nebraska to be picked since there's $4$ left. Or wait.. would it be ${^4C_1}$ which is simply $4$ times ${^6C_2}$ since I only need to pick $1$ person. I basically used the same logic for the $5!$ and $6!$. But I feel like I'm doing this completely wrong. Can anyone correct these for me. 
 A: Your first calculation $(a)$ is correct.
For $(b)$: First, let's ensure that we select $2$ from each state. Order of selection doesn't matter, so we can fill six seats on the committee, two from each state, in the following number of ways:
$$C(6,2)\cdot C(7,2)\cdot C(8,2)$$
That gives us six of seven committee members, with a pool of $21 - 6 = 15$ folks from which we can choose any one, since we already met the condition that each state have two people on the committee, so the last person can be chosen from any of the remaining $15$ people.
So in the end, we have: $$C(6,2)\cdot C(7,2)\cdot C(8,2)\cdot C(15, 1)\;\text{ possible ways}.$$
A: You are counting triple. Please see the linked question.  
Naming the people:
$N = \{N_1, N_2, ..., N_6 \}$
$L = \{L_1, L_2, ..., N_7 \}$
$I = \{I_1, I_2, ..., I_8 \}$  
Let's denote the committee where you choose first $N_1$ and $N_2$ for the two people from Nebrasca and $N_3$ as your $7^{th}$ member as $\{N_1,N_2\}\{N_3\}$.
You are counting the situations $\{N_1,N_2\}\{N_3\}$, $\{N_2,N_3\}\{N_1\}$ and $\{N_3, N_1\}\{N_2\}$ as separate instead of counting them as one $\{N_1,N_2,N_3\}$. Dividing by 3 (the ways you can split 3 elements into a group of 1 and a group of 2), I think, gives the answer.
