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April 4

How do i calculate this special permutation?

I have a permutation that regards two sets to choose from. This will act almost like a combination (order doesnt matter), but in a restricted and unexpected sense. Lets say the sets are as follows:

Set A = {A1, A2, A3, A4) Set B = (B1, B2, B3)

The GENERAL ORDER does matter, but which element is chosen from the same set does not. For instance:

A1, B1, A2, B2, A3, B3, A4 is NOT the same as

A1, B1, A2, B2, A3, A4, B3

But.... A1, B1, A2, B2, A3, B3, A4 IS the same as

A4, B2, A2, B1, A3, B3, A1

.... This is because what matters is the group that is in each position, not the specific element. In other words, you can simplify the permutations to read like this: A B A B A B A, and in cases like that, the elements are indistinguishable. The order still matters to some degree, so this isn't a combinations situation, but how do i eliminate repeats that are counted if i calculate nPr for this special case? It seems hard to talk about when two elements in A are swapped or etc.

.... should i take the permutations of the union of sets a and b, then divide by the permutations of a, and also by permutations of b? This seems reasonable but i am unsure.

Thanks in advance for any assistance! 216.173.144.188 (talk) 18:06, 4 April 2016 (UTC)[reply]

The 4 A's are put in 7 positions in

{\binom {7}{4}}=35

ways. See combination. Bo Jacoby (talk) 18:20, 4 April 2016 (UTC).[reply]

OH! So this IS a combination, but of the "places in the list". In other words, we are taking "7 positions in an ordered list, choose 4", where it doesn't matter what position is picked first, second, etc. What threw me off is i was thinking of the binomial coefficient as choosing ITEMS, not positions. Is this a correct way to see things?

Thanks! 216.173.144.188 (talk) 18:41, 4 April 2016 (UTC)[reply]

Exactly! Bo Jacoby (talk) 23:44, 4 April 2016 (UTC).[reply]

To answer your last question, yes: to count combinations, you count the permutations of the sum and divide by the permutations of the subsets; thus for your problem ${\frac {7!}{4!\,3!}}$ , because the 4! permutations of A and the 3! permutations of B don't matter. —Tamfang (talk) 08:50, 5 April 2016 (UTC)[reply]

Right, i was thinking along those lines at one point Tamfang, but i was unsure of myself. Thanks everyone! :D

216.173.144.188 (talk) 13:05, 5 April 2016 (UTC)[reply]