Prove that the number of permutations of 2n letters consisting of
only a's and b's is greatest when the number of a's is equal to the
number of b's
Let there be m number of "a"s.
Number of permutations = \(\frac {2n!}{m!\times(2n-m)!}\)
Which is equal to 2nCm.
From the property of binomial expansion, we know that middle binomial coefficient is the greatest.
So, m=n makes the value of 2nCm the greatest.
Number of permutations = \(\frac {2n!}{m!\times(2n-m)!}\)
Which is equal to 2nCm.
From the property of binomial expansion, we know that middle binomial coefficient is the greatest.
So, m=n makes the value of 2nCm the greatest.