Question

Question #ff7fe

Answer 1

We divided all `4` letter words into three groups.
Below is the number of combinations and permutations in each group.

Explanation:

Let me interpret the problem in more details.
You would like to use all `11` letters of a word EXAMINATION and choose `4` letters from them at a time.
You are asking how many different `4` letter permutations and combinations are possible.
I hope, that's what you meant.

If all letters in the initial word were different, we would have a textbook problem. Unfortunately, situation gets complicated since three letters, A, I and N are repeated twice.
Let's group all our `4` letter words into three categories:
Group 1. No pairs of the same letters are present in a word.
Group 2. One pair of the same letters is present in a word.
Group 3. Two pairs of the same letters are present in a word.
There can be no more than two pairs present since the total number of letters in a word is `4`.

Group 1
There are `8` different letters in the word EXAMINATION. So, we have `8` candidates for the first place in a word, `7` candidates for the second place, `6` for the third and `5` for the fourth. The total number of words in group 1 (permutations of `4` letters in this group) is
`Perm_(G1) = 8*7*6*5`
The number of combinations of `4` letter in this group is, obviously, smaller by a factor of `4!` since any permutation of these letters gives the same combination:
`Comb_(G1) = (8*7*6*5)/(4!)`

Group 2
There are three choices for a letter that is repeated twice, A, I and N. So, we have `C_3^1` candidates for a pair.
With each of them we have `5` letters to choose for the other two places, which can be done in `C_5^2` ways. So, the total number of combinations of `4` letter in this group is
`Comb_(G2) = C_3^1*C_5^2`
A pair of the same letters can be positioned in a `4` letters word in `C_4^2` ways. With each of them the other two (different) letters can be positioned in `2!` ways. So, the number of permutations in this group is
`Perm_(G2) = C_3^1*C_5^2*C_4^2*2!`

Group 3
Our `4` letters word has two pairs of the same letters. There are three such pairs to choose from, so the number of choices of two pairs is
`Comb_(G3) = C_3^2`
The first pair of the same letters defines an entire permutation. We analyze the number of them for group 2, it's `C_4^2`. So, the number of permutations in this group is
`Perm_(G3) = C_3^2 * C_4^2`