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

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SUMMARY

The discussion centers on calculating the number of groups of 5 dance couples from a pool of 12 males (M) and 10 females (F). The correct method involves using combinations to find the number of groups of 5 males and 5 females separately, then multiplying by the number of ways to arrange those couples, resulting in 23,950,080 combinations. An alternative approach incorrectly assumes that all unique couples can be selected without considering the constraint that each male can only dance with one female at a time, leading to double counting. The correct approach is confirmed to be the first method presented in the discussion.

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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?
 

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