Question #ff7fe

1 Answer
Jun 18, 2016

Answer:

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#