JAMB 2017 Mathematics · Question 27

In how many ways can the word MACICITA be arranged?

Options

  1. A.8!2!\frac{8!}{2!}
  2. B.8!3!2!\frac{8!}{3! 2!}
  3. C.8!2!2!2!\frac{8!}{2! 2! 2!}
  4. D.8!

Answer: C. 8!2!2!2!\frac{8!}{2! 2! 2!}

Explanation written with AI and checked against the answer key

1. Analyze the question

The problem asks for the number of distinct ways to arrange the letters in the word MACICITA. This is a classic problem of permutations with repeated objects.

2. Explain the concept

When arranging a set of nn objects where some objects are identical, we use the formula:

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.
3. Show the solution

Step 1: Count the total number of letters (nn). The word is M-A-C-I-C-I-T-A. Counting them: M(1), A(2), C(3), I(4), C(5), I(6), T(7), A(8). So, n=8n = 8.

Step 2: Identify the frequency of each letter.

  • M: appears 1 time
  • A: appears 2 times (at positions 2 and 8)
  • C: appears 2 times (at positions 3 and 5)
  • I: appears 2 times (at positions 4 and 6)
  • T: appears 1 time

Step 3: Apply the formula. Total arrangements = 8!1!×2!×2!×2!×1!\frac{8!}{1! \times 2! \times 2! \times 2! \times 1!}

Since 1!=11! = 1, we can simplify this to:

8!2!×2!×2!\frac{8!}{2! \times 2! \times 2!}
4. Identify the answer

Comparing our result to the given options: A. 8!2!\frac{8!}{2!} B. 8!3!2!\frac{8!}{3! 2!} C. 8!2!2!2!\frac{8!}{2! 2! 2!} D. 8!

The correct option is C.

5. Exam Tip

Whenever you see a question asking for "arrangements" of a word, always write out the word and count the frequency of each letter immediately. A common mistake is to forget to count one of the repeated letters. Always ensure the sum of the frequencies in your denominator equals the total number of letters in the numerator!

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