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The textbook answer is $\frac{180!} {(5!)^{36}36!}$. This is because the total permutations of 180 items is 180!. The 180! gets divided by $(5!)^{36}$ because the order in each group of 5 doesn't matter. It then gets divided by 36! because the order of the 36 groups doesn't matter.

I calculated this a different way from the textbook. First, I counted groups of 5 from 180 items, which is ${180 \choose 5}=\frac{180!}{5!175!}$ ways. Now from all these different groups of 5, I counted groups of 36, which is ${{180 \choose 5} \choose 36}=\frac{\frac{180!}{5!175!}!}{36!(\frac{180!}{5!175!}-36)!}$.

Is the way that I counted (ie, first count groups of 5, then among those groups, count groups of 36) a valid way to solve this problem? The answer I got was so different from the textbook's I'm not sure how to verify it.

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  • $\begingroup$ You overcount a lot of possibilities by doing this. Your method is not correct. For example, if I end up in group 1 of 5, then in group 36, that's equivalent to me being in group 2 to start, then ending up in group 36 $\endgroup$ – Don Thousand Sep 30 '18 at 22:31
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Yeah, your way of counting doesn't work, for the reason that there will be significant overlaps between the groups counted by $\binom{180}{5}$. If your set is $\lbrace 1, \ldots, 180 \rbrace$, your counting allows for $\lbrace 1, 2, 3, 4, 5\rbrace$ and $\lbrace 1, 2, 3, 4, 6\rbrace$ to be two of your groups.

To follow your reasoning to the answer given in the book, you can count $\binom{180}{5}$ for your first group, then with these $5$ removed (whatever they are), you can count $\binom{175}{5}$ for your second group. Removing these $5$ more ($10$ in total), you can choose one of $\binom{170}{5}$ for your third group, etc, giving you the following count of ordered lists of unordered groups:$\require{cancel}$

\begin{align*} \binom{180}{5} \cdot \binom{175}{5} \cdot \ldots \cdot \binom{10}{5} \cdot \binom{5}{5} &= \frac{180!}{\cancel{175!} \cdot 5!} \cdot \frac{\cancel{175!}}{\cancel{170!} \cdot 5!} \cdot \ldots \cdot \frac{\cancel{10!}}{\cancel{5!} \cdot 5!} \cdot \frac{\cancel{5!}}{0! \cdot 5!} \\ &= \frac{180!}{(5!)^{36}}. \end{align*} However, since we want an unordered group of groups, we have counted each unordered group $36!$ times, yielding a result of $$\frac{180!}{(5!)^{36} \cdot 36!}.$$

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If someone want to use only once the number five and only once the number 36 he has to write the species formula which is:

$E_{36}(E_5[X])$

and means that we have a 36-set whose elements are 5-sets of one sort X.

The associated e.g.f is:

$$\frac{1} {36!} \left ( \frac{x^5}{5!} \right ) ^{36} = {180! \over 36!5!^{36}} {x^{180} \over 180!}$$

and the answer is the coefficient of $ {x^{180} \over 180!}$ i.e. $ {180! \over 36!5!^{36}} $

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