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Say I have a set of 36 objects 0-9A-Z. I want to find out how many different orderings of a dozen selections I can make. This is just 36P12, which is a large number. However, I have two scenarios with their own additional conditions:

  1. The 12 selections MUST include "A" and "H" anywhere in the list. For example, A6ZG2NBQ81H3 and 48HKWP9ACL2J are both valid.

  2. The 12 selections MUST include "A" and "H", but they MUST be seen together as a pair in the list, in "AH" order ("HA" isn't accepted). Additionally, "A" must be in an odd-numbered slot. For example, AH2K86DZYBL5 and 2K86AHDZYBL5 are both valid, but 2AHK86DZYBL5 is not, because "A" is in the second slot.

THE QUESTION: How can I calculate the number of permutations under those two conditions?

MrPuzzler
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Aug 03 '18 at 11:05

1 Answers1

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  1. There are 12 positions (slots) for the A and 11 remaining positions for the H. Then you have to fill the remaining 12-2=10 positions with 36-2=24 characters. Thus: $12 \times 11 \times {34! \over 24!}$.

  2. There are 12/2 = 6 positions for the A, then the location of the H is specified. Then you have 10 remaining positions (slots) to fill with 34 characters. Thus: $6 \times {34! \over 24!}$.

David G. Stork
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