Arranging 14 Coloured Pieces of Paper

Question

I have 3 yellow pieces of paper, 4 blue, 5 brown, and 2 green. I know that I have to do some sort of permutation calculation since a yellow piece in index 1 would be considered the same as a different yellow piece in the same index.

My current solution is : $\dfrac{14!}{3! 4! 5! 2!}=2522520.$ However, I don't know if I've done this correctly.

I was absent for the lesson on how to resolve this and cannot seem to wrap my head around it. Any explanation would be appreciated.

I would say: Permutations with repetition – georg Mar 19 '19 at 02:12 — georg, Mar 19 '19 at 02:12

score 0 · Accepted Answer · answered Mar 19 '19 at 02:12

First assume all 14 papers are considered all different (i.e. papers in same color are indexed with different numbers). Then there are total $$14!$$ cases. Now assume those three yellow papers are considered same. You know there are $3!$ ways to arrange three 'indexed' yellow papers. Now that those $3!$ cases should be merged to 1 case, the total number of cases is now $$\frac{14!}{3!}$$. You do this for four blue papers and now you get $$\frac{14!}{3!4!}$$. You keep doing this to other colors and you get $$\frac{14!}{3!4!5!2!}$$

Thank you for confirming this, Kay. Much appreciated. – Zach Chapman Mar 19 '19 at 02:23 — Zach Chapman, Mar 19 '19 at 02:23

Arranging 14 Coloured Pieces of Paper

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