JAMB 2017 Mathematics · Question 24

In how many ways can the word MATHEMATICS be arranged?

Options

  1. A.11!9!2!\frac{11!}{9!2!}
  2. B.11!9!2!2!\frac{11!}{9!2!2!}
  3. C.11!2!2!2!\frac{11!}{2!2!2!}
  4. D.11!2!2!\frac{11!}{2!2!}

Answer: C. 11!2!2!2!\frac{11!}{2!2!2!}

Explanation written with AI and checked against the answer key

Analysis

To find the number of ways to arrange the letters of a word, we use the formula for permutations of a multiset (arrangements with repeated items). The formula is:

Number of arrangements=n!n1!×n2!×⋯×nk!\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times \dots \times n_k!}

where:

  • nn is the total number of letters.
  • n1,n2,…n_1, n_2, \dots are the frequencies of the repeated letters.
Concept

When some items in a set are identical, swapping them does not create a new unique arrangement. Therefore, we divide the total factorial (n!n!) by the factorial of the count of each repeated letter to remove these redundant arrangements.

Step-by-Step Solution
  1. Count the total number of letters (nn): M-A-T-H-E-M-A-T-I-C-S There are 11 letters in total. So, n=11n = 11.
  1. Identify the frequency of repeated letters: • M appears 2 times. • A appears 2 times. • T appears 2 times. • H appears 1 time. • E appears 1 time. • I appears 1 time. • C appears 1 time. • S appears 1 time.
  1. Apply the formula: Substitute the values into the permutation formula: $Arrangements=11!2!×2!×2!\text{Arrangements} = \frac{11!}{2! \times 2! \times 2!}(Wedivideby (We divide by 2!forM, for M, 2!forA,and for A, and 2!$ for T).
  1. Simplify: The expression is 11!2!2!2!\frac{11!}{2!2!2!}.
Conclusion

Comparing our result to the given options, we can see that it matches option C.

Exam Tip: Always write out the word and tally the frequency of each letter carefully before plugging them into the formula. A common mistake is forgetting to count the letters that appear only once (though they don't affect the denominator since 1!=11! = 1, it helps ensure you haven't missed any letters).

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