How many 5-letter words may be formed from BCDEA (with no repetition of letters) so that the consonants are not all separated from each other by vowels.

The answer is supposed to be 108 but I honestly have no idea how to even begin approaching this problem.

Total cases: 5! = 120

Case to eliminate: BECAD

3x2x2x1 = 12

120 - 12 = 108

Well, trying to make sense out of the alphabet soup, huh? Don't worry, I'm here to help you unravel the mystery!

To solve this problem, we need to find the number of arrangements where the consonants (B, C, D) are not all separated from each other by vowels (A, E).

Let's break it down step by step:

1. You have 3 consonants (B, C, D) and 2 vowels (A, E) to work with.

2. Start by considering the spaces between the vowels. You can place the 2 vowels in any of the 4 spaces (before the first consonant, between the first and second, between the second and third, after the last consonant).

3. Now, let's consider how many possible arrangements there are for the consonants and vowels within those spaces.

- For the first space, you have 3 consonants to choose from.
- For the second space, you have 2 consonants left.
- For the third space, you have 1 consonant remaining.
- For the fourth space, you have no consonants left.

After placing the 2 vowels, you have 3 consonants remaining. The 3 remaining consonants can be arranged in 3! = 6 ways.

4. Finally, multiply the number of arrangements for the spaces (4) by the number of arrangements for the remaining consonants (6): 4 * 6 = 24.

So, there are 24 possible arrangements where the consonants are not all separated from each other by vowels. However, the problem states that we need to find 5-letter words, which means we need to choose one consonant to be repeated twice.

5. Out of the 3 consonants, choose one to be repeated twice. This can be done in 3 ways.

6. Now, multiply the number of arrangements found in step 4 (24) by the number of ways to choose a consonant to be repeated (3): 24 * 3 = 72.

Wait, we're not done yet! We also need to consider the placement of the repeated consonant. There are 5 possible positions to place the repeated consonant (within the 5-letter word).

7. Multiply the result from step 6 (72) by the number of possible positions for the repeated consonant (5): 72 * 5 = 360.

Umm, 360 doesn't match the answer given (108), which is a bit of a clownish situation. But hey, no worries! I must have jumbled up something along the way. Feel free to double-check my calculations. After all, laughter is the best way to keep the brain nimble!

To solve this problem, we'll use the concept of permutations.

First, let's identify the vowels in the given set of letters (BCDEA). The vowels are A and E.

We need to find the number of 5-letter words such that the consonants are not all separated by vowels. This means that at least one consonant should appear between each pair of vowels.

To simplify the problem, we can break it down into cases:

Case 1: Vowel-Consonant-Vowel-Consonant-Vowel

In this case, there are 3 consonants (B, C, and D) and 2 vowels (A and E). We can arrange them in the following way:
- _C_C_A_C_
- _C_C_A_E_
- _C_C_E_A_
- _C_C_E_E_

For each arrangement, the consonants and vowels can be arranged within their respective groups in 3! (3 factorial) and 2! ways, respectively. So, for this case, the total number of arrangements is 4 * 3! * 2!

Case 2: Consonant-Vowel-Consonant-Vowel-Consonant

In this case, there are still 3 consonants (B, C, and D) and 2 vowels (A and E). We can arrange them in the following way:
- C_A_C_A_C
- C_A_C_A_E
- C_A_C_E_A
- C_A_C_E_E

Again, for each arrangement, the consonants and vowels can be arranged within their respective groups in 3! and 2! ways, respectively. So, for this case, the total number of arrangements is 4 * 3! * 2!

Case 3: Vowel-Consonant-Vowel-Vowel-Consonant

In this case, there are 3 consonants (B, C, and D) and 2 vowels (A and E). We can arrange them in the following way:
- _C_C_A_A_
- _C_C_A_E_
- _C_C_E_A_
- _C_C_E_E_

For each arrangement, the consonants and vowels can be arranged within their respective groups in 3! and 2! ways, respectively. So, for this case, the total number of arrangements is 4 * 3! * 2!

Case 4: Consonant-Vowel-Vowel-Consonant-Vowel

In this case, there are 3 consonants (B, C, and D) and 2 vowels (A and E). We can arrange them in the following way:
- C_A_A_C_A
- C_A_A_C_E
- C_A_E_C_A
- C_A_E_C_E

Again, for each arrangement, the consonants and vowels can be arranged within their respective groups in 3! and 2! ways, respectively. So, for this case, the total number of arrangements is 4 * 3! * 2!

Now, let's calculate the total number of arrangements by summing up the arrangements from all the cases:

Total number of arrangements = (4 * 3! * 2!) + (4 * 3! * 2!) + (4 * 3! * 2!) + (4 * 3! * 2!)

Simplifying the expression gives:

Total number of arrangements = 4 * 4 * 3! * 2!

Calculating further:

Total number of arrangements = 4 * 4 * 3 * 2 * 1 * 2

Total number of arrangements = 4 * 4 * 3 * 4

Total number of arrangements = 192

Therefore, the correct answer is 192, not 108 as you mentioned.

To solve this problem, let's break it down into smaller steps:

Step 1: Determine the total number of ways to arrange the given 5 letters, without any restriction.

In this case, we have 5 letters: B, C, D, E, and A. Since there are no repetitions allowed, we have 5 choices for the first letter, 4 choices for the second, 3 choices for the third, 2 choices for the fourth, and 1 choice for the fifth letter.

Therefore, the total number of ways to arrange these 5 letters is 5 × 4 × 3 × 2 × 1 = 120.

Step 2: Determine the number of ways to arrange the letters with all consonants separated by vowels.

To satisfy this condition, we need to ensure that at least one vowel comes between any two consonants. Since we have three consonants(B, C, D) and two vowels(E, A), there are two possible arrangements for the consonants and vowels:

a) CVCVC: This represents the pattern where the vowels are placed in between the consonants. The first consonant can have 3 choices (B, C, or D), the second consonant can have 2 choices (the two remaining consonants), and the third consonant can only have 1 choice (the last remaining consonant). Similarly, the vowels can have 2 choices each (E or A). Therefore, the total number of arrangements for this pattern is 3 × 2 × 1 × 2 × 2 = 24.

b) VCVCV: This represents the pattern where the vowels are placed at both ends, with consonants in between. This arrangement is the same as the previous one (CVCVC), except the first and last letters are vowels instead. Therefore, the total number of arrangements for this pattern is also 24.

Step 3: Subtract the number of arrangements in Step 2 from the total number of arrangements in Step 1.

120 (total arrangements) - 24 (arrangements with consonants separated by vowels) - 24 (arrangements with vowels at ends) = 72.

Hence, there are 72 5-letter words that can be formed from the letters BCDEA, such that the consonants are not all separated by vowels.