This question has two parts:
a) How many anagrams does Mathematics have?
This can be solved by counting the permutations of "mathematics" and removing the permutations of repeated letters. Thus, there are 4,989,600 anagrams of the word "mathematics."$$\frac{11!}{2!*2!*2!*1!*1!*1!*1!*1!} = 4,989,600$$
b) How many 10-letter words can be formed (using only the 26 letters of the alphabet) such that no word is an anagram of another word?
I'm not quite sure how to approach this problem. I know that there are $26^{10}$ possible words of length 10, and I will need to remove permutations of these words to eliminate anagrams. This approach doesn't seem correct though because I feel like I would be over-counting (not sure how to check that either).
If the problem did not allow repeated letters, I think the answer would be ${26\choose10}$ because you would select 10 distinct letters from the available 26. However, since the problem allows repetition, I feel like I might have to use the Principle of Inclusion-Exclusion to subtract all cases in which anagrams occur.
Thoughts?
Best Answer
(a) You've got it. No need to say more.
(b) Short answer: $\binom{35}{10}$.
Long answer: We only care about what letters go into the word, not the order. How many $a$? How many $b$? And so on.
Start at the beginning of the alphabet. We have a choice: either include an $a$, or move on to the next letter $b$. Then we have another choice: include one of the current letter (either $a$ or $b$) or move on to the next letter (either $b$ or $c$). Keep going like this, until we reach $z$ and all ten letters of our "word". We include a letter $10$ times, and move on to the next letter $26-1$ times. There are $\binom{10+26-1}{10}=\binom{35}{10}$ ways to make those choices.
This is often known as the "stars and bars formula".