How many permutations can be formed from objects of types 1, 2, . . . , 2n when objects of type i appear twice when i is odd and objects of type i appear n times when i is even?

Alright so I have no idea on how to tackle or even begin this problem. I've tried thinking of ways to begin, maybe finding the total number of permutations and going from there, but the "i" part of the problem keeps throwing me off.

If you could help me get started that would help, I am a beginner so I might have trouble understanding your hint

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    $\begingroup$ Find out the total number of objects, say $k$. Then find the number of repetitions of objects of the same type. Say for a type, if there are $m$ repetitions, then the answer gets divided by $m!$. You will have to do this for each type. $\endgroup$ – Shubham Johri Apr 7 '19 at 18:29

In general, suppose there are $k$ types of objects, and you have $n_1$ objects of type $1$, $n_2$ objects of type $2$, $\dots$, and $n_k$ objects of type $k$. Then the number of "permutations" of these objects is $$ \frac{(n_1+n_2+\dots+n_k)!}{n_1!n_2!\cdots n_k!}. $$ Here, you have

  • $2$ objects of type $1$,
  • $n$ objects of type $2$,
  • $2$ objects of type $3$,
  • $n$ objects of type $4$,

and so on.

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    $\begingroup$ Yes, that's true @Brownie $\endgroup$ – Shubham Johri Apr 7 '19 at 18:39
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    $\begingroup$ @Brownie Your denominator is correct, but the numerator should be $(2+n+2+n+\dots+2+n)$. To simplify the denominator, count how many times $2$ appears in the denominator, and write it as an exponent, $2^m$. Same for the $n$'s. You can simplify the numerator similarly, since $2+2+\dots+2$ (repeated $m$ times) is equal to $2m$. $\endgroup$ – Mike Earnest Apr 7 '19 at 18:43
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    $\begingroup$ Remember the numerator is not the total number of types, but the total number of objects, which is the sum of the number of objects in each type. $\endgroup$ – Mike Earnest Apr 7 '19 at 18:43
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    $\begingroup$ The total number of types is $2n$. Also, I forgot a factorial in my previous comment; the numerator should be $(2+n+2+n+\dots+2+n)!$. $\endgroup$ – Mike Earnest Apr 7 '19 at 18:52
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    $\begingroup$ @Brownie Don’t forget the factorial in the numerator! Also, each factor in the denominator has a factorial, so it should be $(n!)^n$ down there. Otherwise correct $\endgroup$ – Mike Earnest Apr 7 '19 at 19:14

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