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In forming a committee of 5 out of 5 males and 6 females how many choices you have to made so that there are 3 males and 2 females? How many choices you have to make if there are 2 males?
- a)150
- b)200
- c)1
- d)461
Correct answer is option 'B'. Can you explain this answer?
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In forming a committee of 5 out of 5 males and 6 females how many choi...
Problem Statement:
In forming a committee of 5 out of 5 males and 6 females, how many choices do you have to make so that there are 3 males and 2 females? How many choices do you have to make if there are 2 males?
Solution:
To solve this problem, we use the combination formula which is given as:
nCr = n! / r!(n - r)!
where n is the total number of objects, r is the number of objects chosen, and ! denotes factorial (i.e., the product of all positive integers up to that number).
Part a) - 3 males and 2 females:
To select 3 males out of 5 males, we have 5C3 choices. Similarly, to select 2 females out of 6 females, we have 6C2 choices. Therefore, the total number of ways to form a committee of 5 with 3 males and 2 females is:
5C3 x 6C2 = (5! / 3!2!) x (6! / 2!4!) = (10 x 15) = 150
Therefore, the answer is option (a) 150.
Part b) - 2 males:
To select 2 males out of 5 males, we have 5C2 choices. Similarly, to select 3 females out of 6 females, we have 6C3 choices. Therefore, the total number of ways to form a committee of 5 with 2 males and 3 females is:
5C2 x 6C3 = (5! / 2!3!) x (6! / 3!3!) = (10 x 20) = 200
Therefore, the answer is option (b) 200.
Final Answer:
a) 150
b) 200
In forming a committee of 5 out of 5 males and 6 females, how many choices do you have to make so that there are 3 males and 2 females? How many choices do you have to make if there are 2 males?
Solution:
To solve this problem, we use the combination formula which is given as:
nCr = n! / r!(n - r)!
where n is the total number of objects, r is the number of objects chosen, and ! denotes factorial (i.e., the product of all positive integers up to that number).
Part a) - 3 males and 2 females:
To select 3 males out of 5 males, we have 5C3 choices. Similarly, to select 2 females out of 6 females, we have 6C2 choices. Therefore, the total number of ways to form a committee of 5 with 3 males and 2 females is:
5C3 x 6C2 = (5! / 3!2!) x (6! / 2!4!) = (10 x 15) = 150
Therefore, the answer is option (a) 150.
Part b) - 2 males:
To select 2 males out of 5 males, we have 5C2 choices. Similarly, to select 3 females out of 6 females, we have 6C3 choices. Therefore, the total number of ways to form a committee of 5 with 2 males and 3 females is:
5C2 x 6C3 = (5! / 2!3!) x (6! / 3!3!) = (10 x 20) = 200
Therefore, the answer is option (b) 200.
Final Answer:
a) 150
b) 200
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