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This question already has an answer here:

How many ways are there to form a sequence of 10 letters from 4 a's, 4 b's, 4 c's, and 4 d's if each letter must appear twice?

So typically I would use the theorem like

16! / 4! 4! 4! 4! (if I was choosing a sequence of 16 and exactly 4 each)

For this example though I would have to lock the spots of 2 of each in place so aabbccdd _ _ and those two extra spots would have 4 choices each, so 4 x 4.

Would it be: (10! / 3! 3! 2! 2!) x 16 ?

Any help would be appreciated!

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marked as duplicate by awkward, NCh, Lord Shark the Unknown, JonMark Perry, Parcly Taxel Feb 10 '18 at 7:47

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Your point that you will have aabbccdd__ is right. However note that the two blanks can also be filled by the same letter. Thus you can have 4 a's and two each of the rest. Thus the number of permutations in this case would be 10!/4!*2!*2!*2! . You have 4 letters which are capable of being repeated 4 times . Hence in this case, you can have (10!/4!*2!*2!*2!)*4 permutations.

Next, the 2 blank spaces may be filled in by 2 distinct letters in which case the number of permutations put forth by you 10!/3!*3!*2!*2! would be true. And note that for filling it up with two distinct letters, you have 4*3=12 ways of doing that which gives (10!/3!*3!*2!*2!)*12. Add up the results of these 2 cases to get the required answer.

Note that your answer would be wrong since the number of permutations changes when you have 4 a's and two of each of the rest.

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