Permutation

1) There are 3 pairs of double letter.
The number of arrangements for 11 letters
= 11! / [(2!)(2!)(2!)]
= 4989600

2)
MATHEMATICS
= 3 pairs + 5 single (i.e. MM , T T , AA , H , E , I , C , S)
For 8 letters ,
Case 1 : include 3 pairs + 2 single , total 3C3 * 5C2 types ,
8! / [(2!)(2!)(2!)] arrangements for each types ,
3C3 * 5C2 * 8! / [(2!)(2!)(2!)] = 50400 cases

Case 2 : include 2 pairs + 4 single , total 3C2 * 6C4 types ,
8! / [(2!)(2!)] arrangements for each types ,
3C2 * 6C4 * 8! / [(2!)(2!)] = 453600 cases

Case 3 : include 1 pair + 6 single , total 3C1 * 7C6 types ,
8! / 2! arrangements for each types ,
3C1 * 7C6 * 8! / 2! = 423360 cases

Case 4 : 8 single letters , 1 type ,
8! = 40320 cases

The number of arrangements for 8 letters
= 50400 + 453600 + 423360 + 40320
= 967680