How many $8$ letter words can be formed from the letters $PAAAARRTTM$?

Question

How many $8$ letter words can be formed from the letters $PAAAARRTTM$?

This is what I thought:
Having $8$ letter words is basically just counting the ways of arranging the letters, after we choose to remove $2$ of the letters from the $10$ given.
Thus, I will count the ways of arranging the letters, after I remove some two letters.

The letters I will remove come in $13$ cases and are:
$$\{P,A\},\{P,R\},\{P,T\},\{P,M\},\{A,R\},\{A,T\},\{A,M\},\{A,A\},\{R,T\},\{R,M\},\{R,R\},\{T,M\},\{T,T\}.$$

Removing $P$ and some other letter after:
$$= \frac{8!}{3!2!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!} + \frac{8!}{4!2!2!} = 3780.$$ Similarly for $A$:
$$= \frac{8!}{3!1!2!} + \frac{8!}{3!2!1!} + \frac{8!}{3!2!2!}+ \frac{8!}{2!2!2!} = 13440$$ Similarly for $R$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!2!2!} + \frac{8!}{4!2!} = 2100$$ Similarly for $T$:
$$=\frac{8!}{4!2!2!} + \frac{8!}{4!2!}= 1260$$

So total number of ways is $20580$.
However, the answer is $22260$, with a difference of $1680 = \frac{8!}{4!}$...

Did I go wrong somewhere?
Also, for these type of questions, (given $n$ letters where some repeat and we have to form $k < n$ words from them), is this the systematic way to approach it?

Answer 1

Exponential generating functions provide a systematic way to solve problems like this one.

Let $a_r$ be the number of words of length $r$ that can be formed from the given letters, and let $f(x)$ be the exponential generating function of $a_r$, i.e. $$f(x) = \sum_{r=0}^{\infty}\frac{a_r}{r!} x^r$$ Then $$f(x) = (1+x)^2 \left( 1+x+\frac{1}{2!}x^2 \right)^2 \left( 1+x+\frac{1}{2!}x^2 +\frac{1}{3!}x^3 + \frac{1}{4!}x^4 \right) $$ Expanding $f(x)$, the coefficient of $x^8/8!$ is $22,260$. (I cheated and used a computer algebra system; a less-lazy person could find the coefficient via pencil and paper.)

Answer 2

you made two errors:

for $R$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!} + \frac{8!}{4!2!} $$

for $T$:
$$=\frac{8!}{4!2!} + \frac{8!}{4!2!}$$

$$\text{difference}=(-\frac{8!}{4!2!2!}+\frac{8!}{4!})+(-\frac{8!}{4!2!2!}+\frac{8!}{4!2!})=\frac{8!}{4!}$$

Answer 3

$\newcommand{gtxt}[1]{\bbox[lightgray,4px]{\text{#1}}}$ You can condense this process by considering the possible multiplicity patterns of the 8 selected letters. This avoids the confusion of working through individual combinations.

Here we have the multiset $\{A^4,R^2,T^2,M,P\}$ to choose $8$ items from.

The multiplicity patterns can be written using tokens as $\gtxt{ccccddee}$, $\gtxt{ccccddef}$, $\gtxt{ccccdefg}$, $\gtxt{cccddeef}$, $\gtxt{cccddefg}$, $\gtxt{ccddeefg}$. These are constrained by both the availability of different letters and their multiplicity in the source set.

This pattern also determines how many arrangements there are of each, through the corresponding multinomial, and we also need to calculate how many way there are to fill each pattern. In a pattern with several groups of two, for example, we would pick possible choices to fill those in one step. Then we multiply these two factors to give options for each pattern, and add across all patterns.

$$\begin{array}{|c|c|} \hline \text{For this pattern:} & \text{options to fill} & \text{arrangements} & \text{total options} \\[1ex] \hline \gtxt{ccccddee} & \binom 11 \binom 22 = 1 & \binom{8}{4,2,2} = 420 & 420 \\[1ex] \gtxt{ccccddef} & \binom 11 \binom 21 \binom 32 = 6 & \binom{8}{4,2,1,1} = 840 & 5040 \\[1ex] \gtxt{ccccdefg} & \binom 11 \binom 44 = 1 & \binom{8}{4,1,1,1,1} = 1680 & 1680 \\[1ex] \gtxt{cccddeef} & \binom 11\binom 22\binom 21 = 2 & \binom{8}{3,2,2,1} = 1680 & 3360 \\[1ex] \gtxt{cccddefg} & \binom 11\binom 21 \binom 33= 2 & \binom{8}{3,2,1,1,1} = 3360 & 6720 \\[1ex] \gtxt{ccddeefg} & \binom 33\binom 22 = 1 & \binom{8}{2,2,2,1,1} = 5040 & 5040 \\[1ex]\hline & & \text{total}& 22260 \\[1ex]\hline \hline \end{array}$$

as given.