In how many ways can we arrange the letters of word BAHAMA such that it starts with H and ends with A?

Question

In how many ways can we arrange the letters of word BAHAMA such that it starts with H and ends with A? I have a doubt in the selection of A at the last position. Please help. Thanks in advance!!

Answer 1

In how many ways can we arrange the letters of word BAHAMA such that it starts with H and ends with A?

Method 1: The word BAHAMA has six letters, consisting of three A's, one B, one H, and one M. We must place the H in the first position and an A in the last position, so we have $$H~\square~\square~\square~\square~A$$ That leaves us with two A's, one B, and one M left to place in the four boxes. We choose two of these boxes to fill with the A's, one of the two unused boxes to fill with the B, then place the M in the final unused box. This can be done in $$\binom{4}{2}\binom{2}{1}\binom{1}{1}$$ ways.

I have a doubt in the selection of A at the last position.

The three A's are indistinguishable, so it does not matter which A we place in the last position.

This observation provides us with an alternate approach to the problem.

Method 2: Once the H has been placed in the first position and an A has been placed in the last position, the problem reduces to a permutation of the four remaining letters: A, A, B, M.

Suppose for the moment that the two A's were distinguishable. Let's label them $A_1$ and $A_2$. Then we have four distinct objects to permute: $A_1$, $A_2$, B, M. They can be arranged in $4!$ ways.

However, the two A's are indistinguishable. Hence, we have counted each permutation twice, once when we labeled the first A in the permutation $A_1$ and the second one $A_2$ and once when we labeled the first A in the permutation $A_2$ and the second one $A_1$. For instance, when we replace both $A_1$ and $A_2$ with an A, the two permutations $A_1$B$A_2$M and $A_2$B$A_1$M both become ABAM. Thus, we must divide $4!$ by the $2!$ permutations of the two indistinguishable A's. Thus, the number of distinguishable permutations of the letters A, A, B, M is $$\frac{4!}{2!}$$ As you can check, $$\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{4!}{2!}$$

Answer 2

Place $H$ at the beginning then $A$ at the end. Now permutate the rest $4$ letters and see how many permutation you will have.

There is no double counting as the last latter $A$ is fixed. You can consider it as a special letter $A$ (Denote it by $A^*$ if you want).

Answer 3

All the permutations should start with H and end in A.

That is [H][][][][][A],

$Therefore$ there are 4 blank spaces left and 4 letters (A,A,B,M) left that we arrange in different ways between H and A.

Now number of arrangements will be p(4, 4) because we are arranging 4 things and taking all of them at once.

p(4, 4) = $4!$ = 24 $\cdots$ (1)

but there are 2 copies of a letter (A) in the four letters that we have.

$Therefore$ we divide the result of (1) by 2!

that is 24/2 = 12

$Therefore$ the answer is 12.