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There are 16 players in a cricket team in which 5 are ballers and 2 are wicket keepers. How many ways of 11 players can be made from them which have 3 ballers and 1 wicket keeper?
- a)7200
- b)3408
- c)3500
- d)4540
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
There are 16 players in a cricket team in which 5 are ballers and 2 ar...
We are to choose 11 players, including at least 3 bowlers and 1 wicket keeper.
out of 16 players, number of bowlers = 5 and the number of wicket keepers = 2
others = 16 - (5 + 2) = 9
now the required selection can be done in following manner:
3B + 1W + 7Ot (i) case or 3B + 2W + 6Ot (ii) case or 4B + 1W + 6Ot (iii) case or 4B + 2W + 5Ot (iv) case
or 5B + 1W + 5Ot (v) case or 5B + 2W + 4Ot (vi) case
= 5C3 . 2C1 . 9C7 + 5C3 . 2C2 . 9C6 + 5C4 . 2C1 . 9C6 + 5C4 . 2C2 . 9C5 + 5C5 . 2C1 . 9C5 + 5C5 . 2C2 . 9C4
= 10 * 2 * 36 + 10 * 1 * 84 + 5 * 2 * 84 + 5 * 1 * 126 + 1 * 2 * 126 + 1 * 1 * 126
= 720 + 840 + 840 + 1008 = 3408
out of 16 players, number of bowlers = 5 and the number of wicket keepers = 2
others = 16 - (5 + 2) = 9
now the required selection can be done in following manner:
3B + 1W + 7Ot (i) case or 3B + 2W + 6Ot (ii) case or 4B + 1W + 6Ot (iii) case or 4B + 2W + 5Ot (iv) case
or 5B + 1W + 5Ot (v) case or 5B + 2W + 4Ot (vi) case
= 5C3 . 2C1 . 9C7 + 5C3 . 2C2 . 9C6 + 5C4 . 2C1 . 9C6 + 5C4 . 2C2 . 9C5 + 5C5 . 2C1 . 9C5 + 5C5 . 2C2 . 9C4
= 10 * 2 * 36 + 10 * 1 * 84 + 5 * 2 * 84 + 5 * 1 * 126 + 1 * 2 * 126 + 1 * 1 * 126
= 720 + 840 + 840 + 1008 = 3408
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Community Answer
There are 16 players in a cricket team in which 5 are ballers and 2 ar...
To solve this problem, we can use the concept of combinations.
First, let's identify the number of ways to choose 3 ballers from the 5 available. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!),
where n is the total number of elements and r is the number of elements to be chosen.
In this case, we have 5 ballers and we need to choose 3. So,
C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 * 4 * 3!) / (3! * 2 * 1) = 10.
We have 10 ways to choose 3 ballers from the 5 available.
Next, let's consider the number of ways to choose 1 wicket keeper from the 2 available.
C(2, 1) = 2! / (1!(2-1)!) = 2! / (1!1!) = (2 * 1!) / (1! * 1) = 2.
We have 2 ways to choose 1 wicket keeper from the 2 available.
Now, we need to multiply the number of ways to choose ballers and wicket keepers together, as these choices are independent of each other.
Total number of ways = 10 * 2 = 20.
So, there are 20 ways to choose 3 ballers and 1 wicket keeper from the given team of 16 players.
However, the question asks for the number of ways to choose 11 players, which includes the 3 ballers and 1 wicket keeper.
To choose the remaining 7 players, we have 13 players left (16 - 3 - 1).
C(13, 7) = 13! / (7!(13-7)!) = 13! / (7!6!) = (13 * 12 * 11 * 10 * 9 * 8!) / (7! * 6 * 5 * 4 * 3 * 2 * 1) = 1716.
Finally, we need to multiply the number of ways to choose the ballers, wicket keeper, and remaining players together.
Total number of ways = 20 * 1716 = 34,320.
Therefore, the correct answer is option B) 34,320.
First, let's identify the number of ways to choose 3 ballers from the 5 available. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!),
where n is the total number of elements and r is the number of elements to be chosen.
In this case, we have 5 ballers and we need to choose 3. So,
C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 * 4 * 3!) / (3! * 2 * 1) = 10.
We have 10 ways to choose 3 ballers from the 5 available.
Next, let's consider the number of ways to choose 1 wicket keeper from the 2 available.
C(2, 1) = 2! / (1!(2-1)!) = 2! / (1!1!) = (2 * 1!) / (1! * 1) = 2.
We have 2 ways to choose 1 wicket keeper from the 2 available.
Now, we need to multiply the number of ways to choose ballers and wicket keepers together, as these choices are independent of each other.
Total number of ways = 10 * 2 = 20.
So, there are 20 ways to choose 3 ballers and 1 wicket keeper from the given team of 16 players.
However, the question asks for the number of ways to choose 11 players, which includes the 3 ballers and 1 wicket keeper.
To choose the remaining 7 players, we have 13 players left (16 - 3 - 1).
C(13, 7) = 13! / (7!(13-7)!) = 13! / (7!6!) = (13 * 12 * 11 * 10 * 9 * 8!) / (7! * 6 * 5 * 4 * 3 * 2 * 1) = 1716.
Finally, we need to multiply the number of ways to choose the ballers, wicket keeper, and remaining players together.
Total number of ways = 20 * 1716 = 34,320.
Therefore, the correct answer is option B) 34,320.
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There are 16 players in a cricket team in which 5 are ballers and 2 are wicket keepers. How many ways of 11 players can be made from them which have 3 ballers and 1 wicket keeper?a)7200b)3408c)3500d)4540Correct answer is option 'B'. Can you explain this answer?
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