Question
Question: In a touring cricket team there are 16 players in all including 4 bowlers and 2 wicket keepers. How ...
In a touring cricket team there are 16 players in all including 4 bowlers and 2 wicket keepers. How many teams of 11 players from this can be chosen so as to include three bowlers and 1 wicket keeper?
Solution
In this question we are given a total 16 players out of which 4 are bowlers and 2 are wicket keepers. We need to find the number of ways to form a team of 11 players having 3 bowlers and 1 wicket keeper. For this, we will first find non-bowlers, non-wicket keeper players. Then we will find ways of choosing 3 bowlers out of 4, then 1 wicket keeper out of 2 and at last 7 non-bowler, non-wicket keeper players out of the remaining. We will use a combination method here. Number of ways of selecting r items out of n is given by nCr=r!(n−r)!n!.
Complete step-by-step solution:
Here we are given the total players as 16. Number of bowlers are 4 and number of wicket keepers are 2. Remaining players (non-bowler, non-wicket keeper) will be 16 - (4+2) = 16-6 = 10.
Now we need to form a team of 11 players with 3 bowlers and 1 wicket keeper. So let us find ways of forming such a team.
We know that the total number of ways of selecting r items out of n items is given by nCr=r!(n−r)!n!.
So number of ways of choosing 3 bowlers out of 4 bowlers will be given by 4C3=3!(4−3)!4!=3!×1!4×3!.
Number of ways of choosing 3 bowlers = 4.
Now let us find number of ways of choosing 1 wicket keeper out of 2 wicket keepers we get, 2C1=1!1!2!=2.
Number of ways of choosing 1 wicket keeper = 1.
We are left with 10 players and we have to choose 11 - (3+1) = 7 players out of them.
So number of ways of choosing 7 non-bowler non-wicket keeper out of 10 are:
10C7=7!(10−7)!10!=7!3!10!=7!×3×210×9×8×7!=120.
Therefore total number of ways of selecting a team of 11 players = ways of selecting bowlers × ways of selecting wicket keeper × ways of selecting other players.
4×2×120=960.
Hence the total number of ways of selecting a team becomes 960.
Note: Students should consider every possibility before giving a final answer. Note that we have used multiplication at the end because all these events are occurring simultaneously. Take care while solving nCr=r!(n−r)!n!.