Permutations of 9 total letters, using at least 7 letters. Original Question:

How many different strings can be made from the letters in "EVERGREEN"
  using at least 7 of it's letters.
  Note that the 4 "E"s and 2 "R"s are indistinguishable.

I understand how to use permutations with repetition to determine the number of ways for 9 total letters, but the "at least 7 letters" part stumps me.
Without that condition, it would be 9!/4!2! possible permutations.
Now, should I simply subtract some other number of ways? Perhaps the number of permutations that are less than 7?
How should I proceed? Thanks!
 A: using 9 letters the number of different words is $\binom{9!}{4!2!}$
to find how many words you can form with 8 letters you have to perform the same calculation on each of the following set of letters
EEEVRRGN, EEEERRGN, EEEEVRGN, EEEEVRRN, EEEEVRRG
e.g. for the first one the number of words is $\binom{8!}{3!2!}$, and so on
then you have to add to the previous results the same calculation for 7 letters considering all the different sets, and precisely
EEVRRGN, EEERRGN, EEEVRGN, EEEVRRN, EEEVRRG, EEEERGN,EEEERRN,EEEERRG, EEEEVGN, EEEEVRN,EEEEVRG, EEEEVRR
I don't know whether there exists a different direct way
A: If you use exponential generating functions you can reduce the problem to computing the coefficient of $x^7/7!$ in the expansion of $(1+x)^3(1+x+x^2/2!)(1+x+x^2/2!+x^3/3!+x^4/4!)$.
This is messy but not overwhelming to do by hand if one uses appropriate tricks (eg replace $(1+x+x^2/2!+x^3/3!+x^4/4!)$ with $(x^3/3!+x^4/4!)$ and $(1+x)^3$ with $(3x+3x^2+x^3)$ since the other terms don't contribute to the $x^7$ term)
The most reasonable alternative is to divide up the counting into cases based on the possible partitions that may result from such a set:


*

*Taking 4 E's and 1 each of three out of V,R,G, and N.

*Taking 4 E's, 2 R's, and 1 each of two out of V, G, and N.

*Taking 3 E's, and 1 of each of V,R,G, and N.

*Taking 3 E's, 2 R's, and 1 of each of V,G, and N.

*Taking 2 E's, 2 R's, and 1 of each of V,G, and N.

