Discover the Number of Permutations for 'Examination' | Problem: Permutations

[marutpadhy]

How many permutations of 4 letters can be made out of the letters of the word 'examination'?

 

[soroban]

Hello, marutpadhy!

How many permutations of 4 letters can be made
out of the letters of the word EXAMINATION?

There are 8 distinct letters: A,E,I,M,N,O,T,X.
Three of them are duplicated: AA,\,II,\,NN.

There are three cases to be considered.

[1] Four distinct letters: PQRS
. . ._8P_4 = 1680 pemutations.[2] One pair: PPQR
. . .3 choices for the pair.
. . .Select 2 more letters from the other 7: _7C_2 = 21 ways.
. . .Arrange the 4 letters in {4\choose2,1,1} =6 ways.
There are: .3\cdot 21\cdot6 \:=\:378 permutations.[3] Two pairs: PPQQ
. . .{3\choose2} = 3 choices for the two pairs.
. . .They can be arranged in {4\choose2,2} = 6 ways.
There are: .3\cdot6 \,=\,18 permutations.Total: .1680 + 378 + 178 \:=\:2076 permutations.

 

[marutpadhy]

Thanks for the help. But somehow, I did it ultimately.

But I would also like to mention, that there is some mistake here in the solution:
1. All different: 1068 (fine)
2. One double: 378 (incorrect)
3. Two doubles: 18 (fine)

Now coming to second bit:
What you did there is, you didn't allow the double set to get separated, which is nowhere asked in the problem, just correcting that you get they can be arranged in 12 separate ways.
Just multiply the 3 * 21 * 6 = 756.
Anyways, also this (particularly second bit) could be tried this way:
There are 3 distinct double sets = 3
Remaining 2 spaces can be filled by any two combination of seven other letters: 7!/2!5! = 21
All can be arranged now: *4! = 24(for 4 spaces)
Now, to deal with the duplicacy: /2! (divide by 2 factorial)
Summing all these steps: (3*21*24) / (2) = 756,

Adding all the figures: 1680 + 756 + 18 = 2454.

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By the way, I do not know how to use latex.