I'm struggling to understand the solution to this question concerning combinatorics.
Question:
At a party six boys and six girls dance together. Assuming that classical
dance is performed, in which one couple = one boy and one girl, how many possible
couples can perform together?
My Solution:
There are 6 boys. Each boy can dance with 6 girls. Label the boys $a,b,c,d,e,f$.
Boy $a$ can dance with 6 different girls.
Boy $b$ can dance with 6 different girls.
…
Boy $f$ can dance with 6 different girls.
Therefore, there are $6\times6\times6\times6\times6\times6=6^6=46656$ possible couples. (End of my solution)
The Solution
Fix an enumeration of the six girls and consider all their possible partners.
This is equivalent to permuting the boys, so each girl will dance once with each boy, i.e.,
the total number of possible dancing couples is
$$6!=720$$
(End of solution)
I don't understand:
(i) why my solution is incorrect.
(ii) why the solution provided is correct.
Many thanks in advance.
Best Answer
I think people are failing to address the following issue with you're argument: there is an implicit requirement that all the couples be formed simultaneously. You're thinking about it quite right, except you're forgetting that once a girl has a partner she isn't available for the next boy.