A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10th floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the

Question

A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10th floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

The lift does not stop at 1st and 2nd floors.

Available floors are: 3, 4, 5, 6, 7, 8, 9, 10

Let group 1 have 4 persons and group 2 have 5 persons.

Number of ways in which two groups are formed = $\frac{9!}{4! 5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$ = 126 ways

Number of ways of selecting 2 floors out of 8 floors

Total number of ways in which both groups can leave at different floors = 126 x 28 x 2 = 7056.

Solution

1 2670

2 2840

3 7056

4 11340

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