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I am not 100% sure that I am right here but want to give it a shot.How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
If all the letters were different there would be \(8!\) permutations which end with I, of which \(7!\) start with M, so \(8!-7!=7.7!\) would not start with M but end with I. However there is a repeated A so the final answer is half that.How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
How many permutation of the letters of the words $\bf{"MADHUBANI"}$
do not begin with $\bf{M}$ but end with $\bf{I}$
Don't we need to divide [1] by 2! as well since we consider the cases when the A 's are switched to be the same?
[1] The first letter is $A\!:\;A\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I $
. . .The remaining spaces can be filled in $7!$ ways.
Therefore, there are: .$7! + 6\left(\frac{7!}{2!}\right) \:=\:20,160$ ways.
Both methods are correct, but there is an error in soroban's computation.Don't we need to divide [1] by 2! as well since we consider the cases when the A 's are switched to be the same?
If so, then you'd have \(\displaystyle \frac{7!}{2!}+6\left(\frac{7!}{2!}\right)\) that is \(\displaystyle \left( \frac{7 \cdot 7!}{2!} \right)\) which what CB and I got.
Would like confirmation on this![]()
That red 6 should be a 5: if the first letter is not I, M or either of the two As, then there are only five other choices (count them: D, H, U, B, N). So soroban's count for the number of arrangements should be $7!+5\bigl(\frac{7!}{2!}\bigr)$, which agrees with the answer of Jameson and CaptainBlack.[2] The first letter is not $A$ and not $M\!:$
. . . $\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I$
There are 6 choices for the first letter.
. . . The remaining spaces can be filled in $\frac{7!}{2!}$ ways.