In how many ways we can choose 8 alphabets from aaaaa bbbb ccc dd e

Here's a start:

How many ways can we arrange the a's, b's, c's, d'd and e's?

There are 15 in all, but there are 5 a's, 4 b's, 3 c's, and 2 d's, but only 1 e.

We can arrange them in \frac{15!}{5!4!3!2!1!} ways.

Actually, this is rather tough. I will have to get back to you. If you even care anyway.

This problem requires counting up the 8-permutations of the multiset

{5a, 4b, 3c, 2d, 1e}

Count up the various ways of using these letters and finding an 8-permutation.

Example: aaaaabbb, aaaabbcc, aaaabbbb, and so on and so on. There are quite a few.


Here is an example of an easier one:

Suppose we have S={2a, 1b, 3c}={aa,b,ccc}

Then, acbc, cbcc are 4-permutations of the set S.

The total number of 4-permutations would be

4!(\frac{1}{2!1!1!}+\frac{1}{2!1!1!}+\frac{1}{3!1! }+\frac{1}{3!1!}+\frac{1}{3!1!}+\frac{1}{2!1!1!})

I hope I didn't miss one.

Your problem is much more involved because there are more ways to form 8 permutations out of that multi-set.

Well, when I saw that post, I thought it was a routine one, but then realised the difficulty :o I'm glad you answered it galactus :)

I doubt if there's much point in posting anything else on this topic for the OP's benefit, but I have something interesting to add for those who are.

The number of permutations we have can be found from the generating function

(1+x+x^{2}+x^{3}+x^{4}+x^{5})(1+x+x^{2}+x^{3}+x^{4 })(1+x+x^{2}+x^{3})(1+x+x^{2})(1+x)

If we look at the coefficient of x^8 we see it's 101. That is how many permutations can be made from aaaaa, bbbb, ccc, dd, e in order to sum to 8.

Then, each of those can be arranged in this many ways:

8!\left(\underbrace{\frac{1}{3!3!2!}+\frac{1}{4!3! 1!}+\frac{1}{5!2!1!}+.....................}_{\text {101 of these}}\right)

I wonder why to took it until x raised to the fifth power.. is that because of the 5 'a's?