Let me interpret the problem in more details.
You would like to use all 1111 letters of a word EXAMINATION and choose 44 letters from them at a time.
You are asking how many different 44 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 44 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 44.
Group 1
There are 88 different letters in the word EXAMINATION. So, we have 88 candidates for the first place in a word, 77 candidates for the second place, 66 for the third and 55 for the fourth. The total number of words in group 1 (permutations of 44 letters in this group) is
Perm_(G1) = 8*7*6*5PermG1=8⋅7⋅6⋅5
The number of combinations of 44 letter in this group is, obviously, smaller by a factor of 4!4! since any permutation of these letters gives the same combination:
Comb_(G1) = (8*7*6*5)/(4!)CombG1=8⋅7⋅6⋅54!
Group 2
There are three choices for a letter that is repeated twice, A, I and N. So, we have C_3^1C13 candidates for a pair.
With each of them we have 55 letters to choose for the other two places, which can be done in C_5^2C25 ways. So, the total number of combinations of 44 letter in this group is
Comb_(G2) = C_3^1*C_5^2CombG2=C13⋅C25
A pair of the same letters can be positioned in a 44 letters word in C_4^2C24 ways. With each of them the other two (different) letters can be positioned in 2!2! ways. So, the number of permutations in this group is
Perm_(G2) = C_3^1*C_5^2*C_4^2*2!PermG2=C13⋅C25⋅C24⋅2!
Group 3
Our 44 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_3^2CombG3=C23
The first pair of the same letters defines an entire permutation. We analyze the number of them for group 2, it's C_4^2C24. So, the number of permutations in this group is
Perm_(G3) = C_3^2 * C_4^2PermG3=C23⋅C24
