In a jet there are 3 seats in front and 3 in the back. Number of different ways can six persons of different heights be seated in the jeep, so that every one in front is shorter than the person directly behind is

Options

A. 90
B. 60
C. 54
D. 15

Answer

To solve the problem of seating 6 persons of different heights in a jet with 3 seats in front and 3 seats in the back, where everyone in front must be shorter than the person directly behind them, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Groups**: We have 6 persons and we need to divide them into two groups of 3: one group for the front seats and one for the back seats.

2. **Choose the Front Seats**: We need to choose 3 persons out of the 6 to sit in the front seats. The number of ways to choose 3 persons from 6 is given by the combination formula \( C(n, r) \):
\[ C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = 20 \]

3. **Arrange the Front Seats**: Once we have chosen 3 persons for the front seats, we can arrange them in those seats. Since they must be arranged in increasing order of height (shortest in front), there is only 1 way to arrange them.

4. **Arrange the Back Seats**: The remaining 3 persons will sit in the back seats. Similar to the front seats, they must also be arranged in increasing order of height, which again gives us only 1 way to arrange them.

5. **Calculate Total Arrangements**: The total number of arrangements is the product of the number of ways to choose the front seats and the arrangements of both groups:
\[ \text{Total Arrangements} = C(6, 3) \times 1 \times 1 = 20 \times 1 \times 1 = 20 \]

### Final Answer:
Thus, the number of different ways the six persons can be seated in the jeep, such that everyone in front is shorter than the person directly behind them, is **20**.