arranging letters in Massachusetts

Question

Consider the letters in the word MASSACHUSETTS. Find the number of distinct 13-letter arrangements if the first and last terms must be a vowel. (Leave in factorial form).

By separating into cases where the arrangement started and ended with 2 As, or with an A and E or U, and with E and U, I got $\frac{4 (11!)}{4! 2!}$. However, the answer is apparently $\frac{4\cdot 3 (11!)} {4! 2! 2!}$. What did I miss?

Answer 1

First let us count the number of each letter there are:

$A$-2, $C$-1, $E$-1, $H$-1, $M$-1, $S$-4, $T$-2, $U$-1

For the first and last character to both be vowels, one of three situations will occur:

• Neither $A$ is used as a first or last character

• Exactly one $A$ is used as a first or last character

• Both $A$'s are used as first and last characters

Let us count each case individually.

In the scenario that neither $A$ is used as a first or last character, that implies that the characters are an $E$ and a $U$. Pick which of the two was the first character and the remaining will be the last character. We then choose the locations of the two $A$'s simultaneously and then the four $S$'s and the two $T$'s, or equivalently worded we use multinomial coefficients to find the arrangements of the word aachmsssstt. This gives us: $2\cdot \frac{11!}{2!4!2!}$

In the scenario that exactly one $A$ is used as a first or last character, first select whether the $A$ is the first character or the last character. Then, pick which of the remaining vowels takes the other spot. Then, for whichever vowel is left, we arrange achmssssttx where x is the remaining vowel. This gives us $2\cdot 2\cdot \frac{11!}{4!2!}$

In the scenario that both $A$'s are used as first and last characters, arrange cehmssssttu for the middle. This gives us $\frac{11!}{4!2!}$

Combining all of these together, this gives us a final total of:

$$6\cdot \frac{11!}{4!2!}$$

Their given answer then appears to be correct, though it isn't perfectly clear how they arrived at their answer.

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An alternate approach:

First select the positions for the four vowels collectively. Two of them must be the front and back, so that amounts to selecting two more of the center eleven spaces to be used by vowels. This gives $\binom{11}{2}=\frac{11!}{2!9!}$ options. Then, select the arrangement of the vowels within those four spaces. This can be accomplished in $\frac{4!}{2!1!1!}=4\cdot 3$ ways.

We can then separately arrange all of the consonants in $\frac{9!}{4!2!}$ ways and then thread the two strings together.

This gives us a total of $\frac{11!}{2!9!}4\cdot 3\cdot \frac{9!}{4!2!}=\frac{4\cdot 3\cdot 11!}{4!2!2!}$ ways.

Answer 2

There is no need to consider the different ways it may begin or end in vowels on account of there being two $A$s but only one $E$ or $U$. The standard procedure is to consider multiple instances as individual and overcount, and then at the end divide by the ways to arrange the multiple letters. So although $A_1-----------A_2$ and $A_2-----------A_1$ are the same and count as $1$ whereas $E-----------U$ and $U-----------E$ are different and counts as $2$, we take care of that at the END when we divide by $2!$. That will also be the time we'd take care of, say, $EMA_1SSA_2CHSTTSU$ is the same as $EMA_2SSA_1CHSTTSU$ and only counting as $1$, not $2$. That the $A$s are in the middle, vs. the beginning or end, doesn't make any difference.

So....

Considering the vowels serparately and individually there are $4$ choices for the first letter, then there are $3$ choices for the last. And $11!$ for the rest. So there are $4*3*11!$ ways to arrange it if the individual letters are considered distinct. But they are not. The $4$ $S$s, $2$ $T$s, and $2$ $A$ are considered the same so we must divide by $4!2!2!$.

The answer is $\frac{4*3*11!}{4!2!2!}$

No need to consider separate cases as the dividing by multiple letter arrangements take that into account.

Answer 3

There are four combinations of vowels to start and end the word: $AA, AU, AE, EU$.

If you have anything but $E$ and $U$ at the ends, the permutations for the remaining letters number $11!/(4!2!)$ by multinomial arguments.

If you have $E$ and $U$ at the ends, both $A$'s are in the center, and the permutations for the remaining letters number $11!/(4!2!2!)$.

There are five ways to arrange $AA, AU, EU$ at the front and back, and two ways to arrange $EU$ at front and back.

So, the number of permutations $P$ is

$$P = 5 \cdot \frac{11!}{4!2!} + 2 \cdot \frac{11!}{4!2!2!} = 6 \frac{11!}{4!2!}.$$