Question

If all the letters of the word ‘SENSELESSNESS’ are arranged in all possible ways and an arrangement among them is chosen at random, then, the probability that all the E’s come together in that arrangement is:

• A. \( \frac{1}{990} \)
• B. \( \frac{2}{143} \)
• C. \( \frac{1}{120} \)
• D. \( \frac{1}{429} \)

Hint:

In probability problems involving arrangements, treat repeated items (such as letters) as identical to simplify calculations.

Answer 1

The Correct Option is B

The total number of arrangements of the letters in "SENSELESSNESS" is calculated as: \[ \frac{12!}{3! \times 4! \times 2! \times 2!} = \frac{479001600}{6 \times 24 \times 2 \times 2} = \frac{479001600}{576} = 83160. \] Now, if all the E’s are together, treat the three E's as a single entity. This gives us the arrangement: \[ S, EEE, N, S, L, E, S, S, N. \] Thus, we have 10 entities to arrange. The number of ways to arrange these entities is: \[ \frac{10!}{3! \times 2! \times 2!} = \frac{3628800}{6 \times 2 \times 2} = \frac{3628800}{24} = 151200. \] The probability that all the E's come together is: \[ \frac{151200}{83160} = \frac{1}{143}. \] Thus, the correct answer is \( \frac{2}{143} \).