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.
 A: You have 8 symbols of which 3 are repeats. Of these you are selecting 3 symbols to arrange.
Count the permutations having no, one, two, and three E.
$${5 \choose 3}\frac{3!}{0!} + {5 \choose 2}\frac{3!}{1!}+ {5 \choose 1}\frac{3!}{2!}+{5 \choose 0}\frac{3!}{3!} = 136$$
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If the order of selection is not important, then it is:
$${5 \choose 3} + {5 \choose 2} + {5 \choose 1} + {5 \choose 0} = 56$$
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Case 1: For none of the E, select 3 of the 5 remaining letters. There are ${5\choose 3}$ distinct ways to select these and $3!$ ways to arrange the selected letters.
Case 2: With 1 E, select 2 of the 5 remaining letters.  There are ${5\choose 2}$ ways to select these and $3!$ ways to arrange the selected letters.
Case 3: With 2 selected, select 1 of the 5 remaining letters.  There are ${5\choose 1}$ ways to select these and $\frac{3!}{2!}$ ways to arrange the selected letters.
Case 4: With all of the 3 E, you can select none of the remaining letters.  There are $1$ ways to select these and $1$ way to arrange the selected letters.
This counts all the ways to distinctly select and arrange three of the eight letters in the string "SEQUENCE".
A: 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 $136$.
