SOLUTION: if 4 letters of the word "MATHEMATICAL" are ramdomly selected,how many combinations can be found? How many permutations can be found of the 4 letters selected?

3 of the letters appear more than once in MATHEMATICAL.

There are 3 A's, 2 M's and 2 T's.  1 each of the others.

There are four cases to consider:

Case 1.  No like letters chosen among the four
Case 2.  Exactly 1 pair of like letters chosen among the four
Case 3.  Exactly 2 pairs of like letters chosen among the four
Case 4.  1 triplet of like letters chosen among the four

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Case 1.  No like letters chosen among the four.

Here we can choose them all from this set of 8 letters  {A,C,E,H,I,L,M,T}.

That's 8C4 ways

Case 2.  Exactly 1 pair of like letters chosen among the four.

Here we choose the letter for the 1 pair any of 3 ways, and the other 2 letters
from the other 7.  That's 3*(7C2) ways   

Case 3.  Exactly 2 pairs of like letters chosen among the four.

We can choose the 2 letters for the 2 pairs any of 3C2 ways.

Case 4.  1 triplet of like letters chosen among the four.

We can choose the triplet only one way, as an A, and the remaining 1 letter
from any of the other 7 letters.  That's 7 ways. 

Answer: 8C4 + 3*(7C2) + 3C2 + 7 = 70 + 63 + 3 + 7 = 143

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What we do is to get the permutations of each of
the 4 cases of combinations:


Case 1.  No like letters chosen among the four.
Each one of these is like "MATH".

There are 4! permutations of each of these 
combinations. 

We determined in the first part that
are 8C4 combinations like this.

So the number of permutations of all these 
combinations is:

8C4*4!. 


Case 2.  Exactly 1 pair of like letters chosen 
among the four


Each one is like "MMAT" 

There are 4!/2! permutations of these. (Divide 
by the factorial of the number of each 
indistinguishable letter.)

We determined in the first part that
there are 3*(7C2) combinations like this. 

So the number of permutations of these
combinations is

3*(7C2)*4!/2!.


Case 3.  Exactly 2 pairs of like letters 
chosen among the four.


Each one is like "MMAA" 

There are 4!/(2!2!) permutations of these. 
(Divide by the factorial of the number of 
each indistinguishable letter.)

We determined in the first part that
there were 3C2 combinations like this.
 
So the number of permutations is

3C2*4!/(2!2!)


Case 4.  1 triplet of like letters chosen 
among the four

Each one is like "AAAM"

There are 4!/(3!) permutations of these. 
(Divide by the factorial of the number 
of each indistinguishable letter.)

We determined in the first part that
there are 7 combinations like this.

So the number of permutations is 

7*4!/3!.  

So we add them all up,

8C4*4! + 3*(7C2)*4!/2! 
+ 3C2*4!/(2!2!) + 7*4!/3! = 

70*24 + 3*21*24/2  
+ 3*24/(2*2) + 7*24/6 = 

1680 + 756 + 18 + 28 = 2482 

Edwin