SOLUTION: Show that the number of permutations which can be formed from 2n letters which are either a's or b's is greatest when the number of a's is equal to the number of b's.

Question 549708: Show that the number of permutations which can be formed from 2n letters which are either a's or b's is greatest when the number of a's is equal to the number of b's.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
assume the number of letters is equal to 4.
you can have:
0 a's and 4 b's
1 a and 3 b's
2 a's and 2 b's
3 a's and 1 b
4 a's and 0 b's
when you have 0 a's and 4 b's, the permutation formula is:
4! / (0!*4!) = 1 possible permutations.
that would be:
bbbb
when you have 1 a and 3 b's, the permutation formula is:
4! / (1!*3!) = 4 possible permutations.
that would be:
abbb
babb
bbab
bbba
when you have 2 a's and 2 b's, the permutation formula is:
4! / (2!*2!) = 6 possible permutations.
that would be:
aabb
abab
abba
bbaa
baba
baab
when you have 3 a's and 1 b, the permutation formula is:
4! / (3!*1!)
this is the same as 4! / (1!*3!) which we already did to get you a total of 4 possible permutations.
that would be:
aaab
aaba
abaa
baaa
when you have 4 a's and 0 b's, the permutation formula is:
4! / (4!*0!)
this is the same as 4! / (0!*4!) which we already did to get you a total of 1 possible permutation.
that would be:
aaaa
the formula peaks when the number of a's and b's is equal and goes symmetrically down from there on both sides of the peak.
this is characteristic of the formula.
it works with any number of a's and b's.

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