My Math Forum Permutations

 February 11th, 2007, 07:45 PM #1 Newbie   Joined: Jan 2007 Posts: 18 Thanks: 0 Permutations How many distinguishable permutations of the letters in the word PEPPERY.
 February 12th, 2007, 03:39 AM #2 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 Assuming that you mean permutations that include all the letters, any time you are including permutations where some of the objects are identical, you divide by the number of possible ways to rearrange each group of identical objects. In this case, peppery contains 3 p's, 2 e's, 1 r, and 1y. Notice that these add up to 7, the number of letters in the original word. That is a good way to make sure that you have included all the objects. The number of permutations, then, is 7!/(3!2!1!1!)=420. Make sure you understand this process, and can do it on your own. If you don't, ask more questions until you do.

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