Number of groups of dance couples from pool of M,F

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Homework Statement


How many groups of 5 dances couples can be formed from a pool of {12M, 10F}?

Homework Equations


{}^n\!P_k = \frac{n!}{(n-k)!} \\<br /> {}^n\!C_k = \frac{n!}{k!(n-k)!}

The Attempt at a Solution


We were shown one solution in class which is to find the number of groups of 5M that can be formed from {12M} multiplied by the number of groups of 5F that can be formed from {10F} times the number of groups of 5 couples can be formed from {5M, 5F}:

{}^{12}\!C_5 \cdot {}^{10}\!C_5 \cdot 5! = 23,950,080<br />I thought of alternative approach: find the number of unique couples that can be formed from {12M, 10F} and from that pool find out how many groups of 5 can be formed:

n = 12 \cdot 10 = 120 \\<br /> {}^{120} C_5 = 190,578,024<br />

What is wrong with the 2nd approach?

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The 120 candidate couples includes (June + Dave) and (June + Remy). But at most one of those couples can be in our set of five, as June can only dance with one person at a time. The second method allows two of the five couples to be those two.
 
Ah, thanks. Is there any way to correct for the double counting, or is this approach a non starter?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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