The letters of the word PROMISE are to be arranged in all possible ways, but with the restriction that the three vowels in the word do not all come together as one block. How many different arrangements of the letters of PROMISE are possible under this condition?

Introduction / Context:
This permutation problem involves a restriction on vowel placement. The word PROMISE contains three vowels, and we must count arrangements where these three vowels are not all together in one consecutive block. This type of question is classic in aptitude tests, where it is often easier to count the complement (arrangements with all vowels together) and subtract from the total number of unrestricted arrangements.

Given Data / Assumptions:

• The word is PROMISE.
• Letters: P, R, O, M, I, S, E.
• Total letters: 7, all distinct.
• Vowels: O, I and E (three vowels).
• The condition is that the three vowels must not all come together as one consecutive block.

Concept / Approach:
Let us first count all possible arrangements of the 7 distinct letters without any restriction. Then we count the number of arrangements where the three vowels are together as a single block. Finally, we subtract this latter count from the total to obtain the number of arrangements where the three vowels are not all together. This complement approach simplifies the counting significantly.

Step-by-Step Solution:
Step 1: Count total arrangements without any restriction. Since there are 7 distinct letters, the total is 7! = 5040.Step 2: To count arrangements where all three vowels are together, treat the block of vowels (O, I, E) as one single unit.Step 3: The consonants are P, R, M and S, so along with the vowel block we have 5 units in total: [OIE], P, R, M and S.Step 4: The number of ways to arrange these 5 distinct units is 5!.Step 5: Compute 5! = 120.Step 6: Inside the vowel block, the three vowels O, I and E can be arranged among themselves in 3! ways.Step 7: Compute 3! = 6.Step 8: Total arrangements with all three vowels together equals 5! × 3! = 120 × 6 = 720.Step 9: Therefore, arrangements where the three vowels do not all come together as a single block equals total arrangements minus those with vowels together: 5040 - 720 = 4320.
Verification / Alternative check:
We can check reasonableness: arrangements with all vowels together are a subset of all permutations, so their count must be smaller than 5040. Having 720 such arrangements is plausible, leaving 4320 arrangements where the vowels are not all consecutively grouped. The complement method is standard and reliable for this type of constraint, reinforcing the correctness of the answer.

Why Other Options Are Wrong:
Numbers such as 4694, 4957 and 4871 do not arise from any simple factorial based decomposition of the problem. The total number of unrestricted arrangements is 5040, so no answer larger than this is possible. Among the options, only 4320 equals 5040 minus 720, where 720 is the correct count of arrangements with the vowels together as a block.

Common Pitfalls:
Some students misinterpret the condition and exclude arrangements where any two vowels are adjacent, which is not required. The question only forbids the three vowels from all being together; they can still be partly adjacent. Another common error is to forget to multiply 5! by 3! when counting the arrangements with the vowel block, leading to an incorrect count of 120 instead of 720 and hence an incorrect final answer.

Final Answer:
The number of arrangements of the letters of PROMISE in which the three vowels do not all come together is 4320.