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

Answer: 45