How many combinations can one make with the following 8 letter word? Determine the amount of different 3 character combinations you can form with the characters from SEQUENCE. I imagine the solution is $8*7*6$ but EEE and EEE are not invalid, and EEN and EEN are also the same. I am not sure how to approach this problem.

We separate into **cases**.

Our word will have (a) $3$ E's, or (b) $2$ E's, or (c) $1$ or $0$ E's.

(a) There is only $1$ word of this type.

(b) The _location_ of the E's can be chosen in $\binom{3}{2}$ ways. For each such choice, there are $5$ ways to select the remaining letter, for a total of $15$.

(c) We have $6$ distinct letters available. They can be arranged in $(6)(5)(4)$ ways.

Add up. We get $136$.