How many different 1010-letter permutations can be formed from 88 identical h's and two identical t's?

We are given 8 h's and 2 t's.

A permutation of these letters is simply an arrangement in a row.

For example:

h h h t h h t h h h

or 

h t h h h h h h t h

The number of arrangements is equal to the number of pairs of positions we fix for t, the rest of the positions are filled with h:

these positions are :

    (1, 2), (1, 3), (1,4)                     (1, 10)
           (2, 3), (2,4)                    (2, 10)
                      (3,4)                     (3,10)
 
                                                      (9,10)

so the 1st position combined with any of the remaining 9
the 2 position combined with any of the remaining 8
...
the 9th position combined with the remaining 1

this makes 9+8+7+6+5+4+3+2+1=45 positions to place the 2 t's.

Remark: the number of positions for the 2 t's could also have been calculated by C(10, 2)=10!/(8!2!)=(10*9)/2=45

45

Ithink this answer would be c cause i seen this before

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