Question

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together is:

Hint:

Use casework to handle complex seating arrangements. Consider treating groups as units for easier calculation.

Answer 1

Correct Answer: 17280

To solve the problem of the number of ways 5 boys and 4 girls can sit in a row such that either all the boys sit together or no two boys sit together, we follow these steps:

Case 1: All boys sit together. 

Treat all boys as a single unit. This creates a scenario where we are arranging 1 "boy-block" and 4 girls, which is a total of 5 "units."

The number of ways to arrange these 5 units is given by the factorial of 5:

5! = 120

 

Within the "boy-block," the 5 boys can be arranged among themselves in:

5! = 120

 

Thus, the total number of ways for this case:

120 × 120 = 14400

 

Case 2: No two boys sit together.

Arrange the 4 girls first, occupying 4 positions:

4! = 24

 

Place the 5 boys in the gaps between the girls, which are 5 positions (before the first girl, between each pair of girls, and after the last girl). Choose 5 out of these 5 positions to place boys:

5! = 120

 

Thus, the total number of ways for this case:

24 × 120 = 2880

 

Total number of ways combining both cases:

14400 + 2880 = 17280

 

This calculated value of 17280 falls within the provided range (17280, 17280), confirming its validity.

Answer 2

There are two cases to consider: Case 1: All the boys sit together:
Treat the 5 boys as a single unit. Thus, we have 5 boys (as a block) and 4 girls, which makes 5 units in total. The number of ways to arrange these 5 units is \( 5! \), and within the block of boys, the boys can be arranged in \( 5! \) ways. Hence, the total number of arrangements in this case is: \[ 5! \times 5! = 120 \times 120 = 14400. \] Case 2: No two boys sit together:
Arrange the 4 girls first, which can be done in \( 4! \) ways. This creates 5 possible spaces where the boys can sit. We can place one boy in each of these 5 spaces, and the 5 boys can be arranged in \( 5! \) ways. Thus, the number of arrangements in this case is: \[ 4! \times 5! = 24 \times 120 = 2880. \] Therefore, the total number of arrangements is: \[ 14400 + 2880 = 17280. \]