Does the number of permutations of $n$ objects, $r$ alike of one kind and $n−r$ alike of another kind, always equal the combinations of n different objects taken $r$ at a time? Explain.
I know permutations are $\frac{n!}{(n-r)!}$ and combinations are $\frac{n!}{r!(n-r)!}$
the question is wordy so im not sure what it is asking but is it asking if
$\frac{n!}{(n-r)!}$=$\frac{n!}{r!(n-r)!}$ ? because the answer to that would be no
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$\begingroup$The short answer is: yes
See, you know that the permutation formula for $n$ distinct objects can be simplified to $n!$
But if there are, say $a$ of one kind and $b$ of another, with $a+b=n$, you know that the formula changes to $\frac{n!}{a!b!}$
In the present case, $a= r, b = n-r,$ so formula becomes $\frac{n!}{r!(n-r)!}$, which is identical to the combination formula of choosing $r$ objects at a time from $n$ objects.
Added:
The formula $\frac{n!}{(n-r)!}$ that you mention is the permutation formula of $r$ objects at a time being taken from $n$ objects, i.e. $^nP_r\;\; or\;\; P(n,r) = r!\binom{n}{r}$
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