An engagement team consists of a project manager, team leader, and

C
The number of ways to choose the project manager is 2 and the number of ways to choose a team leader is 3.
The total number of ways to choose 4 consultants out of 7 candidates is 7!/(4!3)= 35, however, we must also take into the consideration that 2 candidates refuse to work together and for that reason, we must deduct from the total number of ways all the options when these 2 are together. We do it by counting all the outcomes, when these 2 are already in team and there are still 2 spots available: 5!/(2!*3!) = 10, so there are 35-10=25 ways of picking 4 candidates out of 7, taking into the consideration the fact that 2 of them don't want to work together. The only thing that remains now is to multiply all outcomes 2*3*25=150

How do we get 5C2 when we calculate the total number of combinations when consultants are together?

Hi TheMastermind ,

When I am saying two consultants are always together, that means I already have them on my bucket and now I have to select the pending 2 from the remaining 5 consultants. Thus, we are using 5C2

Let me know if you need the detailed explanation for this question.

abhimahna Bunuel

I understand the first approach: 2C1 * 3C1 * 7C4 = 2*3*35 = 210

What i donot understand is: Total number of combinations when both consultants are together = 2C1 * 3C1 * 5C2 = 2*3*10=60

I have no idea where the 5C2 comes from instead i have 6C2. If we want to calculate the number of combination that the two consultant are together, we use the Glue methode by compressing the 2 into 1 + the other remaing 5. then we get 6. Then we perform 6C2.

In the end i get this: 210 - (2)(3)(6C2)2!= 210 - (2)(3)(30) = 210 - 180 = 30


Where do i go wrong and where does the 5C2 come from?



Thanks in advance!­

­Select 1 out of 2 PM
Select 1 out of 3 TL
For Consultants there are 7 and 4 to be selected and 2 cant remain in 1 team

Case 1 : One is chosen then total consultant choices 5C3
Case 2: Other is chosen then total consultant choices 5C3
Case 3: None are chosen then total consultant choices 5C4 (both are not selected)

so total choice of consultants = 5C3 + 5C3 + 5C4

Overall choice one after the other = 2C1 X 3C1 X (5C3 + 5C3 + 5C4) = 150