A word "ROORKEE" is given to us. What is the probability that same letters…

Concept: For a word with repeated letters, the number of distinct arrangements of all its letters equals n! divided by the factorial of each letter's repetition count. To force certain letters to always stay together, bundle each such repeated group into one single unit, arrange the reduced set of units, and take probability as (favourable arrangements) / (total arrangements).

Application

1. "ROORKEE" has 7 letters: R, O, O, R, K, E, E — R repeats 2 times, O repeats 2 times, E repeats 2 times, and K occurs once.

2. Total distinct arrangements of all 7 letters = 7! / (2! × 2! × 2!) = 5040 / 8 = 630.

3. For the favourable case, bundle each repeated pair as one block — (RR), (OO), (EE) — together with K, giving 4 distinct units to arrange.

4. Arrange these 4 units: 4! = 24 favourable arrangements (no extra internal arranging is needed inside a block since its two letters are identical).

5. Probability = favourable / total = 24 / 630.

6. Simplify by dividing numerator and denominator by 6: 24 ÷ 6 = 4 and 630 ÷ 6 = 105, giving 4/105.

Cross-check: An independent recomputation confirms the total count differently — choosing positions for the repeated letters via combinations: C(7,2) ways to place R, C(5,2) ways to place O among the remaining slots, C(3,2) ways to place E among what is left, and the last slot for K, giving 21 × 10 × 3 × 1 = 630, matching the earlier total. The two fractions 24/630 and 4/105 are also numerically equal (both ≈ 0.0381), confirming the simplification is correct and no arrangement was double-counted, since dividing by 2!×2!×2! in step 2 already accounts for the identical letters inside each pair.

So the probability that the identical letters stay together is 24/630 = 4/105.