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

Step 1: Understanding the Question
We need to find the total number of ways for 9 people to exit a lift under specific conditions. The conditions are: 1. They exit in two groups, one of 4 people and one of 5 people.
2. The two groups exit at two different floors.
3. The lift only stops at floors 3 through 10.
Step 2: Key Formula or Approach
This is a problem of counting and involves combinations and permutations. The steps are:
1. Determine the number of available floors.
2. Select the people for the first group using combinations \( \binom{n}{r} \). The second group is then automatically formed.
3. Select the two distinct floors where the groups will exit. Since the groups are of different sizes (4 and 5), they are distinct. We can choose an ordered pair of floors. This is a permutation \( P(n,r) \).
4. Combine these steps by multiplication.
Step 3: Detailed Explanation
1. Determine available floors:
The building has floors G, 1, 2, ..., 10. The lift goes up to the 10th floor. People can exit at floors 1 to 10.
The lift does not stop at the first and second floors.
So, the available floors for exiting are \{3, 4, 5, 6, 7, 8, 9, 10\}.
Total number of available floors = 8.
2. Select the groups of people:
We need to divide 9 people into a group of 4 and a group of 5.
The number of ways to choose 4 people out of 9 is given by the combination formula:
\[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \text{ ways} \] Once the group of 4 is chosen, the remaining 5 people automatically form the second group.
3. Assign groups to floors:
The two groups must exit at two different floors, chosen from the 8 available floors.
Let's choose two distinct floors and assign the groups to them. This can be done by choosing an ordered pair of floors. The number of ways to choose 2 floors from 8 and assign the group of 4 to the first chosen floor and the group of 5 to the second is \( P(8,2) \). \[ P(8,2) = \frac{8!}{(8-2)!} = 8 \times 7 = 56 \text{ ways} \] Alternatively, choose 2 floors \( \binom{8}{2}=28 \) and then assign the groups in \(2!\) ways, giving \(28 \times 2 = 56\) ways.
4. Calculate the total number of ways:
The total number of ways is the product of the number of ways to form the groups and the number of ways to assign them to distinct floors.
\[ \text{Total Ways} = (\text{Ways to choose group of 4}) \times (\text{Ways to assign groups to floors}) \] \[ \text{Total Ways} = \binom{9}{4} \times P(8,2) = 126 \times 56 = 7056 \] Step 4: Final Answer
The calculation results in 7056. The correct option based on this calculation is (C).