I'm having a hard time trying to find an intuitive proof of why permutations of n objects including similar objects lets say p, q, r similar objects is $$\frac{n!}{p! q! r!}$$ my question is the same as this question but the answer didn't provide any intuitive explanation
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The linked answer says there are ${n \choose p}$ ways of positioning the first type of objects and for each of those then ${np \choose q}$ ways of positioning the other two. Multiply these together and and expand the binomial coefficients into factorials to get your result, using $r=npq$. This seems fairly intuitive to me – Henry Jan 13 '22 at 14:05

Alternatively, there are $n!$ ways in total of positioning all the objects, but you can independently change the first type positions among themselves $p!$ ways, the second type $q!$ ways and the third type $r!$ ways without changing what you see, so there are $\frac{n!}{p! q! r!}$ distinguishable patterns – Henry Jan 13 '22 at 14:08

1As for seeing why $\binom{n}{p}\binom{np}{q}\binom{npq}{r} = \frac{n!}{p!q!r!}$, expand the binomial coefficients and cancel where applicable: $\binom{n}{p}\binom{np}{q}\binom{npq}{r} = \frac{n!}{p!\color{blue}{(np)!}}\frac{\color{blue}{(np)!}}{q!\color{red}{(npq)!}}\frac{\color{red}{(npq)!}}{r!\color{grey}{(npqr)!}}$, the blues cancel, the reds cancel, and the grey is $0!$ which is equal to $1$ and so can be omitted. For what its worth, the linked answers are all sufficiently intuitive in my opinion. If you would like further clarification, continue in comments there or here. – JMoravitz Jan 13 '22 at 14:11