How many ways to reorder a string's letters If we have to have exactly 7 letters out of which two are "M"s, Two are "X"s and Three are "E"s, without having any consecutive "E"s.
I arrived to the number 10*C(4,2) by just brute forcing the possible positions of the "E"s, but I want to know if there is a standard way to approach this problem. 
 A: You can make the ${4 \choose 2} =6$ possible arrangements of MMXX, then add the E's into 5 possible locations (between characters plus end spaces) in ${5 \choose 3} = 10$ ways for your result of ${4 \choose 2}{5 \choose 3} =60$

When we have "no consecutive" anything or people not sitting together etc, it is often worthwhile leaving those elements until consideration at the end of the process using this kind of method.
A: Joffan wrote an elegant solution.  Here is an alternative method.
To find the number of arrangements in which no E's are consecutive, subtract the number of arrangements in which two or more E's are consecutive from the total number of arrangements. 
We use the Inclusion-Exclusion Principle.  We subtract the number of arrangements in which two consecutive E's appear from the total number of arrangements.  However, this removes the arrangements in which three consecutive E's appear twice (once when the single E appears directly before the block of two E's and once when it appears directly after the block of two E's), so we must add the number of arrangements in which three consecutive E's appear.
The number of arrangements of two M's, two X's, and three E's is the number of ways to fill two of the seven positions with and M, two of the remaining five positions with an X, and the last three positions with the three E's, which is 
$$\binom{7}{2}\binom{5}{2}\binom{3}{3} = \frac{7!}{5!2!} \cdot \frac{5!}{3!2!} \cdot \frac{3!}{3!0!} = \frac{7!}{2!2!3!} = 210$$
The number of arrangements in which two E's are consecutive can be determined by treating the E's as a block.  We now need to arrange six objects, of which two are M's, two are X's, one is EE, and the other is E.  We can do this in 
$$\binom{6}{2}\binom{4}{2}\binom{2}{1}\binom{1}{1} = 180$$
To determine the number of arrangements in which a block of three E's occurs, we treat the three E's as a block, which means we have five objects to place, two M's, two X's, and the block of three E's.  The number of such arrangements is 
$$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 30$$
Hence, the number of arrangements of two M's, two X's, and three E's in which no two E's are consecutive is $$210 - 180 + 30 = 60$$
