Conditional Combinatorics with anagrams I am trying to solve a combinatoric problem with anagrams as follows:
How much anagrams exists from the word "competiras" if they are at least two vowels after the fifth place.
According to the book there are a total of 2,678,400 anagrams. And I try to solve like this:
Solution screenshot

Would appreciate any help.
 A: For the case where there are two vowels after the fifth position, the formula $C(4,2)\cdot 5!^2$ is wrong. You have accounted for choosing which vowels after the fifth position, but you still need to choose which consonants go after the fifth position. Once you have chosen all the letters that go after the fifth position, only then can you permute them in $5!$ ways.
Therefore, the correct count for the first case is
$$
C(4,2)\cdot C(6,3)\cdot 5!^2
$$
I leave it to you to apply this reasoning to the other two cases.
A: The number of permutations with $n$ vowels after the $5$th position:

*

*Choose $n$ vowels $\longrightarrow$ $4 \choose n$ ways.

*The $n$ selected vowels can be ordered in $P(5, n) = n!{5 \choose n}$ ways.

*Choose $5 - n$ consonants $\longrightarrow$ ${6 \choose 5 - n}$ ways.

*Order the $5 - n$ consonants in $(5 - n)!$ ways.

*Order the letters before and including the $5$th position in $5!$ ways.

*Define $T_n =$ total number of permutations $= 5!\cdot n!{4 \choose n}{5 \choose n}{6 \choose 5 - n}(5 - n)!$
Our desired answer hence is
$$\sum_{n = 2}^4 T_n = \sum_{n = 2}^4 \left(5!\cdot n!{4 \choose n}{5 \choose n}{6 \choose 5 - n}(5 - n)!\right) = \boxed{2678400}$$
To get the same answer using overcounting, we have
$$10! - \sum_{n = 0}^1 T_n = 10! - \sum_{n=0}^1 \left(5!\cdot n!{4 \choose n}{5 \choose n}{6 \choose 5 - n}(5 - n)!\right) = \boxed{2678400}$$
