We have 11 letters in "Mathematics"
in which there are
2 M's , 2 T's , 2 A's and other letters H,E,I,C,S are single
Selection of the 4 letters
first case: Two alike and other two alike
In this case we are gonna select the two letters which are alike. We have three choices M,T,A. Out of these, we have to select two(Because we have to select four letters and selecting two alike letters means selecting four letters). So, it can be done in 3C2 ways
second case: Two alike, two different
1 alike letter(which will mean two letters) can be selected in 3C1 ways and other 2 different letters can be selected in 7C2 ways.(as there will be 7 different letters).
So, 3C1*7C2 ways
Third case: All are different
This can be done in 8C4 ways as there are 8 different letters(M,T,A,H,E,I,C,S)
Arrangement:
For the first case, there will be two alike letters. So, arrangement of these letters can be done in 4!/2!*2! ways
For the second case, there will be 1 alike letter and two different letters which can be arranged in 4!/2! ways
For the third case, all letters are different so it can be arranged in 4! ways
So, the final answer will be
(3C2*4!/2!*2!) + (3C1*7C2*4!/2!) + (8C4*4!)
= 18+ 756 + 1680 ways
= 2454 ways