Discussion
25 distinct objects are divided into 5 groups and each group consists of exactly 5 objects. Then the number of ways of forming such groups, is
Answer: C
The number of ways is \(\frac{25!}{(5!)^{6}}\).
First count arrangements into five labelled groups of five: choose 5 for the first, 5 of the rest for the second, and so on.
\[ \begin{aligned} &\binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5} \\ &= \frac{25!}{(5!)^{5}} \end{aligned} \]
The groups are not labelled and all have the same size, so each division has been counted 5! times, once for every ordering of the five groups. Dividing by 5! gives \(\frac{25!}{(5!)^{5}\,5!} = \frac{25!}{(5!)^{6}}\). \(\frac{25!}{(5!)^{5}}\) is the answer for distinguishable groups, which the question does not have.
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