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I have an issue that I thought it would be relatively easy to solve, however it has proven to be tougher than originally thought. I have tried to simplify my problem into the following question(s):

If I have numbers $1-16$ and these numbers have to placed within two groups (say group #$1$ and group #$2$) of $8$ slots each, how do I find the amount of possible permutations where group #$1$ contains numbers $1-8$ and group #$2$ contains numbers $9-16$?

OR

How do I find the amount of permutations in which a number(s) between $9-16$ appears in group #$1$?

Order does matter!

I had originally thought that this could be solved by: $16!-8!8!$, but I have no idea whether this number is correct.

Any help would be greatly appreciated

Jeel Shah
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notmathematician
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1 Answers1

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As I understand the question, there is a certain set, Group $1$, of specified slots into which the numbers $1$ to $8$ are to be placed, and another set, Group $2$, of specified slots into which the numbers $9$ to $16$ are to be placed. Call a permutation that satisfies these conditions good.

The numbers $1$ to $8$ can be permuted within Group $!$ of slots in $8!$ ways. Fpr each of these ways, the numbers $9$ to $16$ can be permuted within Group $2$ of slots in $8!$ ways, for a total of $8!8!$ good permutations.

Now let us count the permutations in which at least one of the numbers $9$ to $16$ appears in Group $1$. These are precisely the permutations which are not good. Since there is a total of $16!$ permutations, the number of not good permutations is $16!-8!8!$, precisely as you had computed.

André Nicolas
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  • Hi Andre, thanks for your input. I'm sorry if this confusing...I guess I did not articulate my problem well. When grouped, I want numbers 1-8 to be in group #1 and 9-16 in group #2. However, I am looking for the amount of possibilities in which this is not the case (i.e. a number outside of 1-8 appears somewhere in group #1). Would this be 8!8! times? – notmathematician Aug 20 '13 at 14:35
  • @notmathematician then just subtract it from the total number of permutations, i.e. $16!-8!8!$. – Bitwise Aug 20 '13 at 14:41
  • Thanks for your help. As you can tell, combinatorics is not my strong point! – notmathematician Aug 20 '13 at 14:46
  • Then you would be looking for the size of the complement and it would be $16!-8!8!$, precisely as you had calculated. – André Nicolas Aug 20 '13 at 14:49