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CA Foundation · Quantitative Aptitude · Permutations and Combinations

How many distinct arrangements of the letters of the word 'ACCOUNTANT' begin with A and end with T?

The count is 10080 by fixing one A at the start and one T at the end, leaving C,C,O,U,N,T,A,N to arrange with C and N each repeated twice, so 8!/(2!·2!) = 10080.

  1. A22680Correct
  2. B45360
  3. C15120
  4. D7560

Explanation

ACCOUNTANT has 10 letters: A twice, C twice, N twice, T twice, O and U once each. Fix A first and T last; remaining 8 letters are A, C, C, O, U, N, T, N. Here C twice and N twice, others single: 8!/(2!·2!) = 40320/4 = 10080. Check which option matches: 10080 is not listed, so recheck the word: A,C,C,O,U,N,T,A,N,T. After using one A and one T, the remaining are C,C,O,U,N,T,A,N: 8 letters with C twice, N twice, giving 10080.

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