To determine the number of permutations of the letters in "PROBLEM" that end in a vowel, let's break down the problem into steps.

Step 1: Identify the Letters and Vowels

The word "PROBLEM" has 7 unique letters: P, R, O, B, L, E, M. Among these letters, there are two vowels, O and E.

Step 2: Restrict the Last Position to a Vowel

Since we want permutations that end in a vowel, the last letter must be either O or E. This gives us two possible choices for the last position.

Step 3: Count Permutations for Each Case

Once we place a vowel at the last position, we have 6 remaining positions to fill with the remaining 6 letters (P, R, B, L, M, and the unused vowel). The number of ways to arrange these 6 letters is: $6! = 720$

Step 4: Multiply by the Number of Vowel Choices

Since there are two choices for the vowel at the last position, we multiply the number of arrangements by 2: $2 \times 720 = 1440$

Conclusion

The total number of permutations of the letters in "PROBLEM" that end in a vowel is: $\boxed{1440}$

Would you like further details, or have any questions?

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Here are 5 related questions to expand your understanding:

1. How many permutations of "PROBLEM" begin with a consonant?
2. How many permutations of "PROBLEM" have the vowels together?
3. How many ways can we arrange the letters of "PROBLEM" with no restrictions?
4. How many permutations of "PROBLEM" have consonants in the first and last positions?
5. How many arrangements of "PROBLEM" can we make if letters can repeat?

Tip: For permutation problems with specific positions, consider fixing those positions first, then calculate the remaining arrangements.