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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 is no female?
- a)150
- b)200
- c)1
- d)461
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
In forming a committee of 5 out of 5 males and 6 females how many choi...
To form a committee of 5 out of 5 males and 6 females:
Total number of choices = 11C5
= (11!/(5!6!))
= (11*10*9*8*7)/(5*4*3*2*1)
= 462
To choose a committee of 3 males and 2 females:
Number of choices = 5C3 × 6C2
= (5!/(3!2!)) × (6!/(2!4!))
= (5*4)/(2*1) × (6*5)/(2*1)
= 10 × 15
= 150
To choose a committee with no females:
Number of choices = 5C5
= 1
Therefore, the correct option is 'C', i.e., 1.
Total number of choices = 11C5
= (11!/(5!6!))
= (11*10*9*8*7)/(5*4*3*2*1)
= 462
To choose a committee of 3 males and 2 females:
Number of choices = 5C3 × 6C2
= (5!/(3!2!)) × (6!/(2!4!))
= (5*4)/(2*1) × (6*5)/(2*1)
= 10 × 15
= 150
To choose a committee with no females:
Number of choices = 5C5
= 1
Therefore, the correct option is 'C', i.e., 1.
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