Question

A lift of a 10 floor building contains 9 persons and group of 4 and 5 leave the lift on different floor and there is no stoppage of lift at 1st and 2nd floor, then find number of ways this can be done.

• A. 7056
• B. 7656
• C. 7066
• D. 7057

Hint:

Always identify the constraints on the "available options" (floors) first. Since the groups exit on "different" floors, we use permutations (\(^n P_r\)) or selection followed by arrangement.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves two parts: first, dividing a group of 9 distinct people into two specific groups of 4 and 5, and second, choosing the floors where these groups will exit. Since the lift does not stop at the 1st and 2nd floors, we must select from the remaining available floors.

Step 2: Key Formula or Approach:
1. Group division: Number of ways to divide \(n\) items into groups of size \(p\) and \(q\) is \(\frac{n!}{p!q!}\).
2. Permutation: Number of ways to assign these groups to floors is \(^n P_r\) where \(n\) is the number of available floors and \(r\) is the number of groups.

Step 3: Detailed Explanation:
1. Total floors available: The building has 10 floors. The lift doesn't stop at floors 1 and 2. Therefore, available floors are 3, 4, 5, 6, 7, 8, 9, 10, which equals 8 floors.
2. Dividing 9 people into groups of 4 and 5: \[ \text{Ways} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \] 3. Selecting 2 different floors out of 8 and assigning the groups: \[ \text{Ways} = ^8 P_2 = 8 \times 7 = 56 \] 4. Total ways: \[ \text{Total} = 126 \times 56 = 7056 \]

Step 4: Final Answer:
The total number of ways the groups can leave the lift is 7056.