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Hardarrangements

~90s

Arranging MATHEMATICS

How many distinct arrangements can be made of the letters of MATHEMATICS?

Show the answer and the worked solution

Correct answer: A.

4,989,600

Worked solution

  1. 11 letters in all, with M, A and T each repeated twice.
  2. Arrangements = 11!/(2! 2! 2!) = 39,916,800 / 8 = 4,989,600.

Answer: 4,989,600.

The idea

Count the letters carefully first, then divide by 2! for each repeated pair. Three pairs means dividing by 8.

Shortcut. Read the options: 39,916,800 is 11! itself and 9,979,200 is 11! ÷ 4. Three repeated pairs need a divisor of 8, so the answer is half of 9,979,200: 4,989,600.

Common trap. "9,979,200" — spotting only two of the repeated pairs (usually M and A, missing the second T) and dividing 11! by 4 instead of 8.

Arranging MATHEMATICS — Aptitude Question

Arranging MATHEMATICS: How many distinct arrangements can be made of the letters of MATHEMATICS? Answer and worked solution on CodeKairo.

Question

How many distinct arrangements can be made of the letters of MATHEMATICS?

Options

  1. 4,989,600
  2. 5,987,520
  3. 9,979,200
  4. 39,916,800
Show the answer and the worked solution

Correct answer: A. 4,989,600

Worked solution

  1. 11 letters in all, with M, A and T each repeated twice.
  2. Arrangements = 11!/(2! 2! 2!) = 39,916,800 / 8 = 4,989,600.

Answer: 4,989,600.

The idea

Count the letters carefully first, then divide by 2! for each repeated pair. Three pairs means dividing by 8.

Shortcut. Read the options: 39,916,800 is 11! itself and 9,979,200 is 11! ÷ 4. Three repeated pairs need a divisor of 8, so the answer is half of 9,979,200: 4,989,600.

Common trap. "9,979,200" — spotting only two of the repeated pairs (usually M and A, missing the second T) and dividing 11! by 4 instead of 8.

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