"The hawaiian" has 11 letters. How many permutations are possible for five of these letters?

This problem is much more difficult. Let's try to solve it.

First, assume temporarily that two "i" letters are different, call them i_1 and i_2. Also call three "a" as a_1, a_2 and a_3, and two h as h_1 and h_2.

Then there are 11*10*9*8*7 = 55440 possible "words" (the first letter is any of 11, the second is any of 10 and so on). But because equal letters actually make the same "words", some...

This problem is much more difficult. Let's try to solve it.

First, assume temporarily that two "i" letters are different, call them i_1 and i_2. Also call three "a" as a_1, a_2 and a_3, and two h as h_1 and h_2.

Then there are 11*10*9*8*7 = 55440 possible "words" (the first letter is any of 11, the second is any of 10 and so on). But because equal letters actually make the same "words", some "words" was counted twice or more times. We have to subtract the number of "parasitic" counts although it is relatively small.

The words that counted more than once are divided into several disjoint sets:
1) with two i's but without repetitions of a and h;
2) with two h's but without repetitions of a and i;
3) with two a's but without repetitions of i and h;
4) with three a's but without repetitions of i and h;
5) with two i's and two a's;
6) with two i's and tree a's;
7) with two i's and two h's
8) with two h's and two a's;
9) with two h's and tree a's.

The first group contains words counted twice and its size is (5*4) * (6*5*4) = 2400 (the first i at one of the 5 places, the second at one of 4, then 11-2i-1h-2a=6 different letters remain). So we have to subtract 2400/2 = 1200. The group 2 also gives -600, and the group 3 too.

The group 4 gives (5*4*3) * (6*5) = 1800 and the words are counted 6 times, -300.

Groups 5, 7, 8: 5*4*3*2*6 = 720 and counted 4 times, so -180.

Group 6 and 9: 5*4*3*2*1 = 120 and counted 12 times, -10.

Totally -(1200*3 + 300 + 180*3 + 10*2) = -4460. So the answer is 55440 - 4460 = 50980.

Uff.