Permutations & Combinations
Arrangements with Identical Objects
Grade None
Question:
<p>INTERMEDIATE has 3 E's, 2 I's and 2 T's and A, D, M, N, R one each, thus total of 12 letters. If words start with I and end with E (i.e. I × × × × × × × × × E), the ten places (shown by cross) have to be filled with 2 E's and 2 T's and 6 distinct letters. Which of the following are correct?</p>
<p>(a) The number of such words is related to arrangements of remaining letters</p>
<p>(b) The total includes arrangements with 2 E's and 2 T's among 10 positions</p>
<p>(c) Some specific count is obtained</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: When a letter is fixed at start and end, we arrange the remaining letters in the middle positions. The count of identical letters changes in the middle portion—we now have 2 E's, 2 T's, and 6 distinct letters (A, D, M, N, R, I) to arrange in 10 positions.
<p><strong>Step 1: Identify the constraint</strong> Word format: I _ _ _ _ _ _ _ _ _ E (with 10 blank positions)</p><p><strong>Step 2: Count remaining letters after fixing endpoints</strong> From INTERMEDIATE's 12 letters, after placing I at start and E at end:</p><ul><li>Remaining E's: 3 - 1 = 2 E's</li><li>Remaining I's: 2 - 1 = 1 I</li><li>Remaining T's: 2 T's</li><li>Other distinct letters: A, D, M, N, R (5 distinct)</li><li>Total to arrange: 2 E's + 1 I + 2 T's + 5 distinct = 10 letters in 10 positions</li></ul><p><strong>Step 3: Apply permutation formula with repetition</strong> Number of arrangements = 10!/(2! × 1! × 2! × 1! × 1! × 1! × 1! × 1!) = 10!/(2! × 2!) = 3,628,800/4 = 907,200</p><p><strong>Step 4: Identify correct options</strong> Based on the calculation of 907,200 arrangements, options A and B are correct (specific option statements would need to be provided in the question to verify exact match)</p><p>∴ Answer: A, B</p>
Correct Answer: A, B