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The problem is this:

  • set A has 12 items
  • set B has 23 items

Taking 4 items from set A and 3 items from set B, what is the number of possible selections?

(Order matters).

I found this: Permutation problem, combining objects from multiple sets

Which would suggest that the answer is P(12,4) x P(23,3) x (4 + 3)!

Is that correct? I'm struggling to understand why the x (4 + 3)! is necessary.

N. F. Taussig
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wetnoodle
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    Order matters is $(4+3)!$ term. – herb steinberg Nov 15 '20 at 19:43
  • Is your problem the same? The $(2+4)$ in the linked answer is used to illustrate why we need $6!$. But in that case they are taking objects from two sets and intermingling the objects which is why we get $6!$. Your problem seems to be different. If I was assigned this I would have left out the $7!$ and left it as the product of permutations because I don't see where we are asked to arrange the selections. – John Douma Nov 15 '20 at 20:10
  • @JohnDouma , I think you are correct! I didn't catch that the linked problem is looking for arrangements. My first instinct was just the product of the permutations but then I was searching to see if I could find similar problems and stumbled upon that one. Thank you! – wetnoodle Nov 15 '20 at 20:20
  • @JohnDouma: But OP has inserted (Order matters) at the end of the question. – true blue anil Nov 16 '20 at 05:06
  • @trueblueanil That's why we use permutations instead of combinations. – John Douma Nov 16 '20 at 05:13

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