MHB Permutations of 'MADHUBANI' Not Starting with 'M', Ending with 'I

[juantheron]

How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$

 

[Jameson]

How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$

I am not 100% sure that I am right here but want to give it a shot.

The starting word has 9 letter with two A's. The number of choices for the first letter is 7 though, because I must be used at the end and M cannot be first, so 9-2=7 choices are left. Now write the number of possible letter choices for the second letter. Again it can't be I but this time it could be M, and we must note that a some letter (not I or M) has been chosen for the first spot. 9-1-1=7 again. Try to continue this process until the end of the word.

Lastly, there are two letters A which are identical so when they switch positions but everything else remains the same, the word is the same and we can't double count. To correct for this error you divide the answer you got above by 2!.

 

[CaptainBlack]

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.

CB

 

[soroban]

Hello, jacks!

I'll give it a try . . .

How many permutation of the letters of the words $\bf{"MADHUBANI"}$
do not begin with $\bf{M}$ but end with $\bf{I}$

We have the letters: .$ A,A,B,D,H,I,M,N,U.$

We want: .$\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I $
n . . . . . . .$\uparrow$
. . . . . . $\sim\!M$[1] The first letter is $A\!:\;A\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I $
. . .The remaining spaces can be filled in $7!$ ways.[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.Therefore, there are: .$7! + 6\left(\frac{7!}{2!}\right) \:=\:20,160$ ways.

 

[Jameson]

[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.

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 [math]\frac{7!}{2!}+6\left(\frac{7!}{2!}\right)[/math] that is [math]\left( \frac{7 \cdot 7!}{2!} \right)[/math] which what CB and I got.

Would like confirmation on this :)

 

[Opalg]

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 [math]\frac{7!}{2!}+6\left(\frac{7!}{2!}\right)[/math] that is [math]\left( \frac{7 \cdot 7!}{2!} \right)[/math] which what CB and I got.

Would like confirmation on this :)

Both methods are correct, but there is an error in soroban's computation.

[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.

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.