Find the number of seating arrangements of eight basketball players on a bench if either the one centre or both of the two forwards must sit at the end where the coach always sits.

case1 : place the centre at the end

number of ways = 1 x 7! = 5040

case2: the two forwards are at the ends
number of ways = 2 x 6! = 1440

total number of ways = 5040+1440 = 6480

To find the number of seating arrangements of eight basketball players on a bench with the given conditions, we can break it down into cases:

Case 1: One centre sits at the end where the coach always sits.

If one centre is already fixed at one end, we can treat the centre as one unit. We have 7 remaining players (including the other centre) to arrange, along with one extra seat.

The number of ways to arrange the 7 players and the extra seat is (7 + 1)! = 8!.

Case 2: Both forwards sit at the end where the coach always sits.

If both forwards are already fixed at one end, we can treat the forwards as one unit. We have 6 remaining players (including the two centres) to arrange, along with one extra seat.

The number of ways to arrange the 6 players and the extra seat is (6 + 1)! = 7!.

Since these two cases are mutually exclusive (only one of them can occur), we need to add the number of arrangements from each case.

Total number of seating arrangements = Number of arrangements in Case 1 + Number of arrangements in Case 2
= 8! + 7!
= 40,320 + 5,040
= 45,360

Therefore, there are 45,360 possible seating arrangements of eight basketball players on a bench if either the one centre or both of the two forwards must sit at the end where the coach always sits.

To find the number of seating arrangements of eight basketball players on a bench under the given conditions, we can break down the problem into separate cases and then add up the number of seating arrangements for each case.

Case 1: Centre at one end, one forward at the other end
In this case, we have 2 options for placing the centre at one end and 2 options for placing the forward at the other end. Once these two positions are fixed, we have 6 players left to arrange in the remaining 6 seats. We can think of this as arranging these 6 players in a line, which can be done in 6! (6 factorial) ways.

So the number of seating arrangements for this case is 2 * 2 * 6!.

Case 2: Centre at one end, both forwards at the other end
In this case, we have 2 options for placing the centre at one end, and 1 option for placing both forwards at the other end. Once these positions are fixed, we have 5 players left to arrange in the remaining 5 seats. Again, we can think of this as arranging these 5 players in a line, which can be done in 5! ways.

So the number of seating arrangements for this case is 2 * 1 * 5!.

To get the total number of seating arrangements, we sum up the number of arrangements from both cases:
Total number of seating arrangements = (2 * 2 * 6!) + (2 * 1 * 5!)

To calculate this, you can use a calculator or perform the factorial calculations manually:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
5! = 5 * 4 * 3 * 2 * 1 = 120

So, the total number of seating arrangements is:
Total number of seating arrangements = (2 * 2 * 720) + (2 * 1 * 120)
= 2880 + 240
= 3120

Therefore, there are 3120 possible seating arrangements for the eight basketball players on the bench given the given conditions.