Difficulty: Easy
Correct Answer: 360
Explanation:
Introduction / Context:
This question tests your knowledge of permutations of multiset letters where some letters repeat. When a word contains repeated letters, we must adjust the standard factorial formula to avoid overcounting identical arrangements that arise from swapping identical letters.
Given Data / Assumptions:
Concept / Approach:
If all letters were distinct, the number of arrangements would be 6! because there would be 6 positions and 6 distinct choices for each ordering. However, here E is repeated twice. Interchanging the two E letters does not create a new distinct word. To correct for overcounting, we divide by the factorial of the number of times each repeated letter appears. Thus the formula becomes total permutations = 6! / 2! for this word.
Step-by-Step Solution:
Total letters n = 6.If all were distinct, arrangements = 6! = 720.Letter E is repeated 2 times, so divide by 2! to account for identical swaps.Compute 2! = 2.Hence distinct arrangements = 6! / 2! = 720 / 2 = 360.
Verification / Alternative check:
There are no other repeated letters, so only one division factor (2!) is needed.You could manually check a smaller example such as a three letter word with two identical letters to see that division by 2! is correct in principle.
Why Other Options Are Wrong:
420 and 576 would require non integer division by a repetition factor or some incorrect counting logic.220 is too small and does not match any simple factorial division for this word.
Common Pitfalls:
Forgetting to divide by 2! for the repeated letter E, which would give 720 and double the correct answer.Dividing by an extra factorial such as 3! by incorrectly assuming another repeated letter.Confusing combinations with permutations and using 6C2 or similar expressions.
Final Answer:
The letters of LEADER can be arranged in 360 distinct ways.
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