9. A touring party of 20 cricketers consists of 9 batsmen, 8 bowlers and 3 wicket-keepers. A team of 11 players must have at least 5 batmen, 4 bowlers and 1 wicket-keeper. How many different teams can be selected:

a) If all the players are available for selection
b) If 2 batsmen and 1 bowler are injured and cannot play

Not being familiar with the game of cricket,

I find this sentence confusing:
"A team of 11 players must have at least 5 batmen, 4 bowlers and 1 wicket-keeper"

Assuming that you would want 11 players, that could mean:
6 batmen, 4 bowlers, 1 wicket-keeper;
5 batmen, 5 bowlers, 1 wicket-keeper;
5 batmen, 5 bowlers, 2 wicket-keepers

Bob's calculation would be for precisely
5 batmen, 4 bowlers, and 1 keeper, but that is only 10 players
So I think we must include:
C(9,6)*C(8,4)*C(3,1) + C(9,5)*C(8,5)*C(3,1)+C(9,5)*C(8,4)*C(3,2)

number:

9!/(5!4!) * 8!/(4!4!) * 3!/2!1!

b. 7!/(5!2!)*7!/(4!3!)*3!/2!1!

See the pattern?

I agree, I often can't count past 10.

a) If all the players are available for selection, we can calculate the number of different teams by considering the required number of players in each category.

Number of ways to select 5 batsmen out of 9 = C(9, 5) = 126
Number of ways to select 4 bowlers out of 8 = C(8, 4) = 70
Number of ways to select 1 wicket-keeper out of 3 = C(3, 1) = 3

Total number of different teams = (number of ways to select batsmen) * (number of ways to select bowlers) * (number of ways to select wicket-keepers)
= 126 * 70 * 3
= 26,460

So, if all the players are available for selection, there can be 26,460 different teams.

b) If 2 batsmen and 1 bowler are injured and cannot play, we need to recalculate the number of different teams considering the reduced number of available players.

Number of ways to select 3 batsmen out of 7 (9 - 2) = C(7, 3) = 35
Number of ways to select 3 bowlers out of 7 (8 - 1) = C(7, 3) = 35
Number of ways to select 1 wicket-keeper out of 3 = C(3, 1) = 3

Total number of different teams = (number of ways to select batsmen) * (number of ways to select bowlers) * (number of ways to select wicket-keepers)
= 35 * 35 * 3
= 3,675

So, if 2 batsmen and 1 bowler are injured and cannot play, there can be 3,675 different teams. However, I hope they have good backups!

To find the number of different teams that can be selected, we can use combination formulas.

a) If all the players are available for selection, we need to select 11 players from a total of 20 players in the touring party. We have the constraint that we must have at least 5 batsmen, 4 bowlers, and 1 wicket-keeper.

We can break down the selection process into selecting the required number of batsmen, bowlers, and wicket-keepers, and then selecting the remaining players from the remaining pool of players.

The number of ways to select the batsmen is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of batsmen (9) and k is the required number of batsmen (5 or more). So, the number of ways to select 5 or more batsmen is:
C(9, 5) + C(9, 6) + C(9, 7) + C(9, 8) + C(9, 9) = 126 + 84 + 36 + 9 + 1 = 256.

Similarly, the number of ways to select the bowlers is: C(8, 4) = 70.
And the number of ways to select the wicket-keeper is: C(3, 1) = 3.

Finally, we need to select the remaining 2 players from the pool of remaining players (which includes batsmen, bowlers, and wicket-keepers).

The number of ways to select these 2 players is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of remaining players (20 - number of batsmen - number of bowlers - number of wicket-keepers) and k is the number of players to be selected (2). So, the number of ways to select the remaining 2 players is:
C(20-9-8-3, 2) = C(20-20, 2) = C(0, 2) = 0.

Therefore, the total number of different teams that can be selected is the product of all these possibilities:
256 * 70 * 3 * 0 = 0.
So, if all players are available, there are no different teams that can be selected because there are not enough players to meet the requirements.

b) If 2 batsmen and 1 bowler are injured and cannot play, then we now have restrictions on the selection process.

We need to select 11 players from a total of 20 players in the touring party, but we must now select the required number of batsmen, bowlers, and wicket-keepers from the available players. We can calculate the number of ways to select the batsmen, bowlers, and wicket-keepers, and then select the remaining players from the pool of remaining players.

The number of ways to select the batsmen is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of batsmen remaining after the injured players are removed (9 - 2 = 7) and k is the required number of batsmen (5 or more). So, the number of ways to select 5 or more batsmen is:
C(7, 5) + C(7, 6) + C(7, 7) = 21 + 7 + 1 = 29.

The number of ways to select the bowlers is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of bowlers remaining after the injured player is removed (8 - 1 = 7) and k is the required number of bowlers (4 or more). So, the number of ways to select 4 or more bowlers is:
C(7, 4) + C(7, 5) + C(7, 6) + C(7, 7) = 35 + 21 + 7 + 1 = 64.

The number of ways to select the wicket-keeper is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of wicket-keepers remaining after the injured player is removed (3 - 1 = 2) and k is the required number of wicket-keepers (1). So, the number of ways to select 1 wicket-keeper is:
C(2, 1) = 2.

Finally, we need to select the remaining 3 players from the pool of remaining players (which includes batsmen, bowlers, and wicket-keepers).

The number of ways to select these 3 players is calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!), where n is the total number of remaining players (20 - number of batsmen - number of bowlers - number of wicket-keepers - number of remaining injured players) and k is the number of players to be selected (3). So, the number of ways to select the remaining 3 players is:
C(20-7-7-2-0, 3) = C(4, 3) = 4.

Therefore, the total number of different teams that can be selected is the product of all these possibilities:
29 * 64 * 2 * 4 = 14848.
So, if 2 batsmen and 1 bowler are injured and cannot play, there are 14,848 different teams that can be selected.