A project team has to be formed such that 1 out of 3 candidates is to

Question

A project team has to be formed such that 1 out of 3 candidates is to be selected as director, 2 out of 4 candidates are to be selected as managers, and 3 out of 8 candidates are to be selected as project experts. If a particular manager needs to be selected and 2 particular project experts refuse to be in the same team, then how many different project teams can be formed?

A. 400
B. 450
C. 500
D. 550
E. 600

A. 400
B. 450
C. 500
D. 550
E. 600

Kinshook · Answer

Given: A project team has to be formed such that 1 out of 3 candidates is to be selected as director, 2 out of 4 candidates are to be selected as managers, and 3 out of 8 candidates are to be selected as project experts.

Asked: If a particular manager needs to be selected and 2 particular project experts refuse to be in the same team, then how many different project teams can be formed?

Number of ways to select a director out of 3 candidates = 3C1 = 3
Number of ways to select 2 managers out of 4 candidates when a particular manager needs to be selected = (4-1)C(2-1) = 3C1 = 3
Number of ways to select 3 project experts out of 8 candidates when 2 particular project experts refuse to be in the same team = 6C3 + 2C1*6C2 = 20 + 2*15 = 50
Number of ways to select project team = 3C1*3C1*(6C3+2C1*6C2) = 3*3*(20+2*15) = 9*50 = 450

IMO B

nahid78 · Answer

A director can be selected in 3 ways (3c1)
A manager is already selected so we have to select another manager in 3 ways. (one out of remaining three)
Now we can select project experts by Total ways of selecting 3 experts out of 8 MINUS selecting such way that both the particular experts are selected.
 on other words 8c3 - (How both of them can be selected)
 = 8c3 - 6c1 (AS both of them are already selected we can select the third expert from remaining 6 people.
 = 56 - 6
 = 50

So answer is 3*3*50
=450

nahid78 · Answer

Another way of selecting 3 people out 8 such a way that particular two experts are not selected together is whether select 3 people from the six expert or select one from the 2 expert and another two from the other 6.
So, 6c3 or 2c1*6c2
Total = 6c3 + 2c1*6c2 = 20 + 2*15= 50

ramjas1234 · Answer

Could you please explain the selection of experts here. managers and directors are fine.
But for experts, why 6C3 i.e 20 isn't the correct answer, since 2 out of 8 refused, hence 6 left and 3 to be selected out of those 6, that's it. where is 2C1*6C2 coming from and why?

Bunuel · Answer

Because “2 particular project experts refuse to be in the same team” doesn’t mean both are out, it only means they can’t be on the team  together .

So we split:

Both excluded: 6C3 = 20 
 Exactly one included: 2C1 * 6C2 = 30 
 Total = 50

Alternatively, you can do it as total minus restriction:

Total ways without constraint: 8C3 = 56 
 Forbidden ways (both included): choose both (2C2) * choose 1 more from the remaining 6 (6C1) = 6 
 So 56 - 6 = 50.

Hope it's clear.

AdarshSambare · Answer

Number of ways to select a director = C(3,1) = 3
Number of ways to select 2 managers when a particular manager must be selected = C(3,1) = 3
Number of valid ways to select 3 project experts out of 8 when 2 particular experts refuse to be together:
Total C(8,3) = 56; subtract those containing both (choose both × choose 1 of remaining 6 = 6) → 56 − 6 = 50
Total number of teams = 3 × 3 × 50 = 450
Answer: B

A project team has to be formed such that 1 out of 3 candidates is to

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