Question 465983: Can someone answer this question for me step by step or in a broken down manner
Determine the number of permutations of the letters of the word "sedlection."
thanks....I saw it answered as 9![2!]=181,440 i was just wonder how you came up with this answer i looked at similar questions and determined that there are 1s,2e's,1L,1c,1t,1i,1o,and 1n...but after this i am lost what do you do from here thanks
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! "Selection" not "sedlection."
There are 9! ways to arrange nine distinct letters. However we divide by 2! because of the E's. If we were to treat the E's as distinct there would be 9! ways but here, we can arrange the E's themselves in 2! ways, leaving 9!/2! = 181440.
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