General

@Blackyoshi: Oh... I probably learned this and forgot, sorry :<

http://imgur.com/LCW0LhU,6Ue89Me#0
http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

Here, I did some digging; and was able to secure the process

[quote=Neuro]this won't work because you're not guaranteed to use the repeating lettters[/quote]

mb, im retarded

yeah just go case by case

it's basically what 1st poster said, i just wrote it out.
be careful to add ALL possible cases. it's easy to slip up and miss one or more cases, and end up with a wrong answer =p

[quote=bloodIsShed]First, count each letter:
E = 4, V = 1, R = 2, G = 1, N = 1
^for reference

consider all string with
2 E, 2 R, 1V, 1 G, 1 N => 7! / (2! 2!)
3 E, 1 R, 1 V, 1G, 1N => 7! / 3!
3 E, 2 R, 1 V, 1G => 7! / (3! 2!)
3 E, 2 R, 1 V, 1N => 7! / (3! 2!)
3 E, 2 R, 1 G, 1N => 7! / (3! 2!)
4 E, 2 R, 1 V => 7! / (4! 2!)
4 E, 2 R, 1 G => 7! / (4! 2!)
4 E, 2 R, 1 N => 7! / (4! 2!)
4E, 1R, 1V, 1G => 7! / 4!
4E, 1R, 1V, 1N => 7! / 4!
4E, 1R, 1G, 1N => 7! / 4!
4 E, 0 R, 1 V, 1G, 1N => 7! / (4!)

(notice that any other combination of E and R will not result in a string with 7 characters)

So sum is [7! / (2! 2!)] + [7! / 3!] + 3 [7! / (3! 2!)] + 3 [7! / (4! 2!)] + 4[7! / 4!]
1260 + 840 + 3*420 + 3* 105 + 840
= 4515

edit: fixed again[/quote]

Ohhhhhhh ok I get it now! Thank you.

[quote=Blackyoshi]How come I can't exclude 2 letters which aren't E or R?[/quote]

Oh ya, I forgot about that...

[quote=FaTaLP3NGU1N]Find the cases where you exclude two Rs, two Es, one R, one E or 1 R and 1 E

Case 1: Exclude 2 Es
Remaining letters: VRGREEN
Amount of strings: 7!/2!2!

Case 2: Exclude 2 Rs
Remaining letters: EVEGEEN
Amount of strings: 7!/4!

Case 3: Exclude 1 R and one letter which is not E (I'll use the exclusion of V as a general case)
Remaining letters: EEGREEN
Amount of strings: 7!/4!
but you can exclude G or N as well, so there are three times of these results
Total amount of strings for this case: 3(7!/4!)

Case 4: Exclude 1 E and one letter which is not R (I'll use the exclusion of V as a general case)
Remaining letters: ERGREEN
Amount of strings: 7!/3!2!
Again, you can choose to exclude G or N as well, so there are three of this result too
Total amount of strings for this case: 3(7!/3!2!)

Case 5: Exclude 1 E and 1 R
Remaining letters: EVGREEN
Amount of strings: 7!/3!

Sum of all cases = 4200[/quote]

How come I can't exclude 2 letters which aren't E or R?

First, count each letter:
E = 4, V = 1, R = 2, G = 1, N = 1
^for reference

consider all string with
2 E, 2 R, 1V, 1 G, 1 N => 7! / (2! 2!)
3 E, 1 R, 1 V, 1G, 1N => 7! / 3!
3 E, 2 R, 1 V, 1G => 7! / (3! 2!)
3 E, 2 R, 1 V, 1N => 7! / (3! 2!)
3 E, 2 R, 1 G, 1N => 7! / (3! 2!)
4 E, 2 R, 1 V => 7! / (4! 2!)
4 E, 2 R, 1 G => 7! / (4! 2!)
4 E, 2 R, 1 N => 7! / (4! 2!)
4E, 1R, 1V, 1G => 7! / 4!
4E, 1R, 1V, 1N => 7! / 4!
4E, 1R, 1G, 1N => 7! / 4!
4 E, 0 R, 1 V, 1G, 1N => 7! / (4!)

(notice that any other combination of E and R will not result in a string with 7 characters)

So sum is [7! / (2! 2!)] + [7! / 3!] + 3 [7! / (3! 2!)] + 3 [7! / (4! 2!)] + 4[7! / 4!]
1260 + 840 + 3*420 + 3* 105 + 840
= 4515

edit: fixed again

Find the cases where you exclude two Rs, two Es, one R, one E or 1 R and 1 E

Case 1: Exclude 2 Es
Remaining letters: VRGREEN
Amount of strings: 7!/2!2!

Case 2: Exclude 2 Rs
Remaining letters: EVEGEEN
Amount of strings: 7!/4!

Case 3: Exclude 1 R and one letter which is not E (I'll use the exclusion of V as a general case)
Remaining letters: EEGREEN
Amount of strings: 7!/4!
but you can exclude G or N as well, so there are three times of these results
Total amount of strings for this case: 3(7!/4!)

Case 4: Exclude 1 E and one letter which is not R (I'll use the exclusion of V as a general case)
Remaining letters: ERGREEN
Amount of strings: 7!/3!2!
Again, you can choose to exclude G or N as well, so there are three of this result too
Total amount of strings for this case: 3(7!/3!2!)

Case 5: Exclude 1 E and 1 R
Remaining letters: EVGREEN
Amount of strings: 7!/3!

Sum of all cases = 4200

[quote=MeMagicalPie]EVERGREEN has 9 letters
E is repeated 4 times and R is repeated 2 times

9! / 4! x 2![/quote]

but I'm supposed to choose only 7 of the 9 letters... D:

EVERGREEN has 9 letters
E is repeated 4 times and R is repeated 2 times

9! / 4! x 2!

[quote=Blackyoshi]Oh so would it be 3C2 7!/(4!2!) ?[/quote]

There would never be a case where you make a string without E therefore,
case 1: 1 E
case 2: 2 Es
case 3: 3 Es
case 4: 4 Es

since there are 9 letters avaliable, u can chose to exclude the Rs
case 5: 0 Rs
case 6: 1 R
case 7: 2 Rs

edit: there will be more cases

[quote=Neuro]u have to find all the cases then add them up[/quote]

Oh so would it be 3C2 7!/(4!2!) ?

[quote=gamemage3]divide out the permutations of the repeating letters
9*8*7*6*5*4*3/4!2![/quote]

this won't work because you're not guaranteed to use the repeating lettters

divide out the permutations of the repeating letters
9*8*7*6*5*4*3/4!2!

u have to find all the cases then add them up