Difficulty: Easy
Correct Answer: 5040
Explanation:
Introduction / Context:
This question is a straightforward permutation problem where all letters in a word are distinct. We are asked to count the number of different sequences that can be formed using all letters of the word POVERTY, without any additional conditions.
Given Data / Assumptions:
Concept / Approach:
When all letters are distinct and we use all of them, the number of arrangements equals the factorial of the number of letters. Since there are 7 distinct letters, the count of permutations is 7!.
Step-by-Step Solution:
Step 1: Count the number of distinct letters in POVERTY. There are 7 distinct letters.Step 2: To form an arrangement using all letters, we are essentially permuting 7 distinct items.Step 3: The number of permutations of 7 distinct items is 7!.Step 4: Compute 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1.Step 5: Calculate step by step: 7 * 6 = 42, 42 * 5 = 210, 210 * 4 = 840, 840 * 3 = 2520, 2520 * 2 = 5040.Step 6: Therefore, there are 5040 distinct arrangements.
Verification / Alternative check:
Why Other Options Are Wrong:
Common Pitfalls:
Students sometimes mistakenly think some letters repeat, for example confusing POVERTY with words that have double letters, and use formulas for permutations with repetitions. Another pitfall is arithmetic error when computing 7!. Writing out the factorial multiplication step by step and double checking helps avoid such mistakes.
Final Answer:
The number of different arrangements of the letters of POVERTY is 5040.
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