How many ways can you arrange the letters from 'GREEN' with at least one 'E'?

[hms.tech]

Homework Statement

In how many ways can the three letters from the word " GREEN " be arranged in a row if atleast one of the letters is "E"

Homework Equations

Permutations Formula

The Attempt at a Solution

The total arrangements without restriction: 5P3/2! = $\frac{5!}{2! * 2!}$

The number of arrangements in which there is no "E" = 3!

Ans : 5P3/2! - 3! = 24 (wrong)

Here is another approach :

The arrangements with just one "E" = $\frac{2!*4!}{2!}$
The arrangements with two "E" = $\frac{3*2}{1}$

I think I making a mistake due to the repetition of "E" ... Can anyone of you tell me a better way which avoids the problem I am being having .

 

[HallsofIvy]

If "at least one letter must be E", that means that the other two letters must be chosen from "GREN". That is, the first letter must be one of those 4 letters, the second one of the three remaining letters. How many is that?

But then we can put the "E" that we took out into any of three places: before the two, between them, or after the two letters so we need 3 times that previous number.

 

[hms.tech]

If "at least one letter must be E", that means that the other two letters must be chosen from "GREN". That is, the first letter must be one of those 4 letters, the second one of the three remaining letters. How many is that?

But then we can put the "E" that we took out into any of three places: before the two, between them, or after the two letters so we need 3 times that previous number.

4P2 = 4*3

According to your method the answer should be 12 * 3 = 36 (The correct answer in the solutions is "27")

Clearly your method (as did mine) repeats some of the permutations :

You didn't take onto account that the two "E" are not distinct. Thus in those permutations where we chose two "E" were repeated. see :
GEE , EEG, EGE
ENE, EEN, NEE
REE, ERE , EER

 

[CAF123]

4P2 = 4*3

According to your method the answer should be 12 * 3 = 36 (The correct answer in the solutions is "27")

Clearly your method (as did mine) repeats some of the permutations :

You didn't take onto account that the two "E" are not distinct. Thus in those permutations where we chose two "E" were repeated. see :
GEE , EEG, EGE
ENE, EEN, NEE
REE, ERE , EER

Alternatively, split the problem into two: Count those combinations with only 1 E and then count separately those with two E's.

For one E combination: (1C1)*(3C2)*3! = 18
For two E combination: (2C2)*(3C1)*(3!/2!) = 9. Then add.