Of 5 employees, 3 are to be assigned an office and 2 are to

Well, your method is good, but I think this is a more safe method:

The committee should look like this : Man, Man, Woman

how to choose 3 people to be assigned in office from 5: 5C3 = 5!/3!2! = 10

How to choose 2 man and 1 woman= 3C2 * 2C1 = 3!/2! * 2!/1! = 6

Probability = Number of Desired outcome /Total possible outcome = 6/10 = 3/5

offices can be allocated as m,m,w or m,w,m or w,m,m
P(m,m,w) = 3/5*2/4*2/3
P(m,w,m) = 3/5*2/4*2/3
P(w,m,m) = 2/5*3/4*2/4

Since its 'OR', need to add all the 3 probabilities. So answer is 1/5+1/5+1/5 = 3/5
somehow am not comfortable with combinations, hence chose to take a direct method.

Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?

the question is asking for the probability of SELECTING 2 men & 1 woman, it does not matter if the selection includes woman A OR B.
hope it helps.

Hi oss198,
Sorry, I saw your post today only.

Arrangement is out of context here. We are finding the ways of selecting the 3 persons out of 5. (either you choose ABC or BCA from ABCDE is same here)

Selection of 3 persons from 5 :- 5 C 3
selection of 3 persons from 5 and arranging them on 3 chairs :- 5 C 3 * 3!
Selection of 5 persons from 5 :- 5 C 5
Selection of 5 persons from 5 and arranging them on 5 chairs :- 5 C 5 * 5! --------> which ultimately turns out to be 5!

If you need brief insights of these concepts check my article, Permutations and Combinations (In my Signature).

Hope that helps!

Thanks a lot for your answers Aldossari and Narenn.
Narenn your signature is awesome i'll have a look.

1) 3/5*2/4*2/3=2/10 (where 3/5 probability of 1 man being selected, 2/4 the other man, 2/3 one woman)
2) 2/10*3=3/5 (multiplying by 3 as there are 3 cases in different order)

When 3 of the 5 employees are assigned an office, the remaining 2 employees automatically will be assigned a cubicle. Therefore, the number of ways to choose 3 employees to be assigned an office is 5C3 = 5!/3!(5-3)! = (5 x 4 x 3)/(3 x 2 x 1) = 10.

Now we need to find the number of ways in which 2 of the 3 employees selected to have an office will be male and 1 of them will be female. Since there are 3 male employees and 2 female employees, the number of ways to choose 2 male employees out of 3 males and 1 female employee out of 2 females is 3C2 x 2C1 = 3 x 2 = 6.

Finally, since there are 10 ways to choose 3 employees from 5, and 6 ways to choose 2 male employees from 3 and 1 female employee from 2, the probability is 6/10 = 3/5.

Answer: D

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