# Question ff7fe

Jun 18, 2016

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
$P e r {m}_{G 1} = 8 \cdot 7 \cdot 6 \cdot 5$
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
$C o m {b}_{G 2} = {C}_{3}^{1} \cdot {C}_{5}^{2}$
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
$C o m {b}_{G 3} = {C}_{3}^{2}$
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
$P e r {m}_{G 3} = {C}_{3}^{2} \cdot {C}_{4}^{2}$