Distinct arrangements of the letters in BANANA

Question

Consider all the ways to arrange the 6 letters of the word 	ext{BANANA}. Two arrangements count as the same if they read as the same string (the three A's are interchangeable, and so are the two N's). How many distinct arrangements are there?

Accepted Answer

If all 6 letters were different, there would be 6! = 720 orderings. But some of those look identical because letters repeat. BANANA has one B, three A's, and two N's. Any single arrangement is counted 3! times over from shuffling the three A's among themselves, and 2! times from swapping the two N's. To remove this overcount, divide: \frac{6!}{3!\,2!\,1!} = \frac{720}{6 	imes 2 	imes 1} = \frac{720}{12} = 60. Takeaway. Arrangements of a word with repeated letters equal \dfrac{n!}{n1!\,n2!\cdots}, dividing by the factorial of each repeat count. BANANA gives 60.

Distinct arrangements of the letters in BANANA

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