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**B**

- This is a combination-probability problem. To solve this we will need to find out the number of ways we can choose 2 men out of 6 and 3 women out of 4 to sit in the chair.
- Next, we note that the probability of seating two men and three women in the empty chairs is:

(number of ways of choosing 2 men and 3 women)/ (number of ways of seating five people out of ten). - The total number of ways of choosing five people to sit out of a total of ten (six men + four women) is given by:

_{10}C_{5}= 10!/([10-5]!5!)

= (10*9*8*7*6)/[10-5]!

= (10*9*8*7*6)/5!

= 252 - Now we want the seats to be taken by 2 men and 3 women.

I. The number of ways to choose 2 men out of six to occupy the seat is given by:

_{6}C_{2}= 6!/4!*2! = 15

II. The number of ways to choose 3 women out of 4 to sit is given by:

_{4}C_{3}= 4!/3!*1! = 4 - Using the multiplication principle, the total number of ways to select two men and three women is:

15*4=60 - The probability is thus = 60/252

= (12*5)/(12*21)

= 5/21. Hence the correct answer choice is B

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