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 :

• A. \(2184\)
• B. \(3064\)
• C. \(7056\)
• D. \(11340\)

Answer 1

The Correct Option is D

Concept: If \(n\) persons are to be divided into two groups of sizes \(r\) and \(n-r\), the number of ways is \( \binom{n}{r} \). If two groups exit on different floors, we also multiply by the number of ways to choose those floors. Step 1: {Determine the possible floors where the lift can stop.} The building has floors \(1\) to \(10\). The lift does not stop at floors \(1\) and \(2\). Thus possible exit floors: \[ 3,4,5,6,7,8,9,10 \] Total \(=8\) floors. Step 2: {Choose two different floors for exit.} Since two groups exit at different floors: \[ \binom{8}{2} = 28 \] Step 3: {Choose which 4 persons exit at the first chosen floor.} \[ \binom{9}{4} = 126 \] Remaining \(5\) persons exit at the other floor. Step 4: {Assign the two groups to the chosen floors.} Two groups can be arranged in \(2!\) ways. \[ 2! = 2 \] Step 5: {Compute total number of ways.} \[ 126 \times 28 \times 2 \] \[ = 7056 \] Thus the number of ways is \(7056\).