A word made by writing the seven letters used to spell "EXAMPLE" in some order. Find how many different 'words' are possible in each of the following cases.

a) The first and the last letter are each "E"
b) The two letters "E" are next to each other.
c) The two letters "E" are not next to each other.

Do you need to use all the letters for the new word?

For example, b) PEEL

a) If the first and last letter have to be "E," then we only need to arrange the remaining 5 letters. There are 5! (5 factorial) ways to do this, which equals 120. So, we have 120 different "words."

b) If the two letters "E" have to be next to each other, we can consider them as a single entity. That means we now have 6 entities to arrange: EE, X, A, M, P, and L. We can think of the two Es as "twins." So, the number of different arrangements is 6!. However, within the EE entity itself, the two E's can switch places without changing the word. So, we need to divide by 2 to eliminate duplicates. Therefore, we have 6! / 2 = 360 / 2 = 180 different "words."

c) If the two letters "E" are not next to each other, we can treat them as separate entities. This means we have 7 entities to arrange: E, X, A, M, P, L, and E. Since "E" appears twice, we need to divide by 2 to account for the duplicate arrangements within the E's. So, we have 7! / 2 = 5040 / 2 = 2520 different "words."

Just remember, these are all different "words," not necessarily real words. So, don't expect to find all of them in the dictionary!

a) If the first and the last letter are each "E", we have to arrange the remaining 5 letters (X, A, M, P, L) in between. The word will have the following pattern: E _ _ _ _ _ E.

To calculate the number of different arrangements, we need to consider that each letter is distinct, so we use the formula for permutations of distinct objects.

The number of different arrangements is 5!, which is equal to 5 x 4 x 3 x 2 x 1 = 120.

b) If the two letters "E" are next to each other, we can consider them as a single entity. So, we have X, A, M, P, L, and the combination EE. This gives us the pattern: _ _ _ _ _.

Now, we need to calculate the number of different arrangements of these 6 distinct objects. Again, we use the formula for permutations of distinct objects.

The number of different arrangements is 6!, which is equal to 6 x 5 x 4 x 3 x 2 x 1 = 720.

c) If the two letters "E" are not next to each other, we have two cases to consider.

Case 1: E _ _ _ _ E. In this case, we have the remaining letters X, A, M, P, L. We need to calculate the number of different arrangements, which is the same as in part a), giving us 120 possibilities.

Case 2: _ E _ _ _ E _. In this case, we have the remaining letters X, A, M, P, L. We need to arrange the 5 distinct objects, which is the same as in part a), giving us another 120 possibilities.

Adding the number of possibilities from both cases, we get a total of 120 + 120 = 240 different words.

To solve this problem, we need to consider the seven letters used to spell "EXAMPLE" and find the number of different words that can be formed based on the given conditions.

a) The first and the last letter are each "E":
In this case, the first and last letter of the word must be "E". So the two "E" letters are fixed in their positions. Now we need to arrange the remaining five letters (X, A, M, P, L) among the three remaining positions. This can be done in 5P3 ways (permutation of 5 letters taken 3 at a time).
Therefore, the number of different words possible when the first and last letter are each "E" is 5P3 = 5! / (5-3)! = 5! / 2! = 5 * 4 * 3 = 60.

b) The two letters "E" are next to each other:
In this case, we treat the two "E" letters as a single entity. So we have six elements (EXAMPL) and need to arrange them among the remaining two positions (excluding the first and last position, which are occupied by the two "E" letters). This can be done in 6P2 ways (permutation of 6 letters taken 2 at a time).
Therefore, the number of different words possible when the two "E" letters are next to each other is 6P2 = 6! / (6-2)! = 6! / 4! = 6 * 5 = 30.

c) The two letters "E" are not next to each other:
In this case, we have to arrange the seven letters (E, X, A, M, P, L, E) in such a way that the two "E" letters are not adjacent. To solve this, we subtract the number of cases where the two "E" letters are next to each other (found in part b) from the total number of possible arrangements of the seven letters (which is 7P7 = 7!).
Therefore, the number of different words possible when the two "E" letters are not next to each other is 7P7 - 6P2 = 7! - 6! / (7-2)! = 7! - 6! = 5! * 7 - 6! = 5280.

In summary,
a) The number of different words possible when the first and last letter are each "E" is 60.
b) The number of different words possible when the two "E" letters are next to each other is 30.
c) The number of different words possible when the two "E" letters are not next to each other is 5280.