number:
9!/(5!4!) * 8!/(4!4!) * 3!/2!1!
b. 7!/(5!2!)*7!/(4!3!)*3!/2!1!
See the pattern?
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
3 answers
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)
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)
I agree, I often can't count past 10.