# Please explain logic behind a given solution

#### X-man

##### New Member
The following example is from Ryser's book Combinatorial Mathematics:

"The number of 12-letter words that can be formed from 4 A's, 4 B's, 2 C's and 2 D's is 12!/(4!4!2!2!) = 207,900."

I assume the reasoning behind this is that the 12 letters available can be permuted in 12! ways, but the 4 A's are indistinguishable as are the 4 B's, 2 C's and 2 D's, and that's why we divide by the product of 4!, 4!, 2!, and 2!.

Right after that Ryser gives this example: "The number of 5-letter words that can be formed from the letters A, B, C in which A appears at most twice, B at most once, and C at most three times is

5!/(2!0!3!) + 5!/(2!1!2!) + 5!/(1!1!3!) = 60."

I don't understand Ryser's solution. I solve it following the logic of the first example given above. I reason:

1) Since A appears at most twice, B at most once, and C at most three times, we're really asking: how many 5-letter words can we form from two A's, one B, and three C's = six letters?

2) Thus we count the number of permutations of five letters taken from the six letters, which is 6!/( 6–5 )! = 6!

3) But since the two A's are indistinguishable as are the three C's, we have to divide this by 2!3! to get

6!/(2!3!) = 60.

Can someone explain the reasoning that leads to Ryser's answer, which is a sum of three addends?

#### BGM

##### TS Contributor
1) Since A appears at most twice, B at most once, and C at most three times, we're really asking: how many 5-letter words can we form from two A's, one B, and three C's = six letters?
Not really; The question is still asking for a 5 letter permutation. But now the question gives 1 extra letter of "freedom" so you have 3 possible scenarios

$$\{(2,1,2), (2,0,3), (1,1,3)\}$$

under the constraint $$a \leq 2, b \leq 1, c \leq 3, a + b + c = 5, a, b, c \in \mathbb{Z}^+\cup\{0\}$$

#### X-man

##### New Member
Hi, BGM,

The "not really" distinction you make is too subtle for my brain. Can you formulate a problem where Ryser's approach and mine would result in different answers?

But thanks for the reply. I'd thought Ryser might be using multinomial coefficients, but couldn't quite work that out.

#### Dason

##### Ambassador to the humans
Consider making a 2 letter word with at most 2 As, 1 B, and 1C.

Your method says that we look at the number of permutations of 2 letters out of the 4: 4!/2! = 12 then divide by 2! since there are 2 As. So we get an answer of 6.

But if you work through it the alternative way you get an answer of 7 and indeed these are the possibilities:
"aa" "ac" "ca" "ab" "ba" "bc" "cb"

#### X-man

##### New Member
EXCELLENT, DASON! Your post will give me something to think about and work on over the eggnog!