In how many ways can the word MATHEMATICS be arranged?

  • A 11!/(9! 2!)
  • B 11!/(9! 2! 2!)
  • C 11!/(2! 2! 2!)
  • D 11!/(2! 2!)
  • E \(\frac{11!}{2! \cdot 2! \cdot 2!}\)

The correct answer is C. 11!/(2! 2! 2!)

The word "MATHEMATICS" has 11 letters, but it has repeated letters: 2 "M"s, 2 "A"s, and 2 "T"s. To find the number of ways the letters can be arranged, we use the concept of permutations with repetition.

The formula for arranging \(n\) items with repetition, where \(n_1, n_2, \ldots, n_k\) are the counts of repetitions of each item, is:

\(\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}\)

In this case, \(n = 11\) (total letters) and we have repetitions of \(M\), \(A\), and \(T\), so \(n_1 = 2\), \(n_2 = 2\), and \(n_3 = 2\).

Substitute the values into the formula:

\(\text{Number of ways} = \frac{11!}{2! \cdot 2! \cdot 2!}\)

Now, calculate the value:

\(\text{Number of ways} = \frac{11 \cdot 10 \cdot 9 \cdot \ldots \cdot 3 \cdot 2 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{11!}{2! \cdot 2! \cdot 2!}.\)

Therefore, the correct answer is:

B. \(\frac{11!}{2! \cdot 2! \cdot 2!}\)

Previous question Next question