An engagement team consists of a project manager, team leade

We can either choose 4 consultants from a group of 5 (5C4) who are willing to work with everyone or choose 3 from that group (5C3) and one person from the other group of 2 consultants (2C1) who don't want to work with each other.

I get C. If I see this on the test, id prolly get to 210 and then guess A B or C.

OK so for the first position we have only 2 possiblities ( proj manager) 3 for the team leader so and 7!/3!4! for the consultants

2*3*35 --> 210

Now I dunno how to figure out the constraints quickly but I eventually figured it out. Obvs. we want to figure out the total ways in which the two ARE on the same team.

I did it by AB are the two and XYZFN are the rest

ABXO (O stands for the other 4) So the first is ABXY, Z,F,N 4 possible choices

The next is

ABYO (but notice we already had YX so there are only 3 possible choices)
ABZO
ABFO
ABNO No possible here we used them all up

So 4+3+2+1 = 10

SO now its 2*3*10 =60

So 210-60 = 150

2C1*3C1*(5C4+5C3*2C1)=2*3*(5+10*2)=6*25=150 -> C

Can you explain this part is words:

(5C4+5C3*2C1)

Answer: C
a) No of ways to select 1 Manager = 2c1 = 2
b) No of ways to select 1 Team leader = 3c1 = 3
c) No of ways to select 4 Consultants = 7c4 = 35
Therefore, possible teams without any constraint = 2x3x35 = 210

No of ways to select 4 Consultants out of 7 when 2 of them are always together = 6c4 x2! = 60

Therefore, possible teams with given constraint = 210 - 60 = 150

Hope its clear.

Well, probably the quickest way to do this problem is to

1) eliminate D & E, because we are pretty sure it is less than 210 (we know max is 2*3*35).
2) eliminate A & B, because we know 25 and 35 are not multiple of 6 (we know the combination will be an integer).

Simple and straightforward! (Also, I used this method as well! )

I'd like to add that your B is actually
2C2 * 5C2 = 1 * 10 = 10
We have 2C2, because we have already chosen the consultants in the group.

Hi All,

The answer choices to this question provide us with an interesting 'shortcut' that we can use to avoid some of the math involved.

Since there are 2 possible project managers and 3 potential team leaders, then the final answer MUST be a multiple of (2)(3) = 6.

Since we're choosing 4 of 7 possible consultants, we can use the Combination Formula:

7!/(4!3!) = 35 possible groups of 4 consultants.

IF there were no additional restrictions, then there would be (6)(35) = 210 possible groups. HOWEVER, we know that certain consultants won't work with other consultants, so the number of possible groups must be LESS than 210. Based on the answer choices, there's only one option that is a multiple of 6 and is less than 210...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

i did this question in this way
2c1 * 3c1* 5c4+5c3*2c1= 2*3*(5+5*4*2/2)=6*25=150

We can select the project manager in 2 ways and the team leader in 3 ways. For the consultants, there are 7 candidates for 4 positions, but 2 of the 7 cannot be together.

We can use the following equation:

# of ways to select the consultants = # of ways with the 2 together + # of ways with the 2 not together.

The total number of ways to select the consultants is 7C4 = 7!/[4!(7-4)!] = 7!/(4!3!) = (7 x 6 x 5 x 4)/4! = 7 x 6 x 5 x 4)/(4 x 3 x 2) = 35.

The number of ways to select the consultants when the 2 are together can be found by choosing 2 people for the remaining slots from the 5 remaining candidates, which is given by 5C2 = (5 x 4)/2 = 10.

Thus, the number of ways to select the consultants when the 2 are not together is 35 - 10 = 25.

So, the total number of ways to select the teams is 2 x 3 x 25 = 150.

Answer: C

Given a team is to be formed of 1 PM, 1 TL & 4 consultants, out of 2 candidates for the position of PM, 3 candidates for TL & 7 for consultants.

Constraint is that 2 of the 7 consultants cannot be on same team. Let these 2 be called A & B

We have two cases:

Case 1: Choose one of the 2 consultants ( A or B) & put him on the team & choose the rest 3 from the remaining 5.

We get # of ways as = 2C1 * 5C3 * 2C1 * 3C1 = 2 * (5*4*3/1*2*3) * 2 *3 = 120

Case 2: Exclude both A & B and choose the 4 consultants from the remaining 5 consultants.

We get # of ways as = 5C4 * 2C1 * 3C1 = 5 * 2 * 3 = 30

Hence total # of ways = 120 + 30 = 150

Answer C.


Thanks,
GyM

Hi! Can someone please explain what is wrong with this approach :
2c1 * 3c1 * 5c3 * 2c1 ( 5c3 = 5 without issues consultants for 3 positions and 2c1 = 2 consultants who have problem have only spot)

Hi kunalbean,

Your calculation assumes that 1 of the 4 "consultant" spots must be taken by 1 of the 2 candidates who will not work with one another. However, it's possible that NEITHER of those 2 candidates are selected for those 4 spots - so you have to account for those additional options.

GMAT assassins aren't born, they're made,
Rich

Thank you for the response! One quick question - As you've got 800 - how can one ensure of thinking through all the possibilities ? for example, in this case, the one you stated above.

Hi kunalbean,

Consistent and proper note-taking is one of the 'keys' to maximizing your Score on Test Day, so when you're facing a situation in which you have to think about more than one possible outcome, you would almost certainly find it helpful to take some quick notes on what outcomes are. Sometimes the simple act of writing out ideas that you already know can get you thinking about other aspects of the question (especially those that you might not have considered).

For example, in this prompt, you might write down:

PM: 2 possibilities
TM: 3 possibilites
Consultants: 7 possible (A,B,C,D,E,F,G), choose 4 - A and B CANNOT be on the team together.

That last bit is especially important: it does NOT state "choose A or B"... it states "those 2 can't be on the team together." There's nothing in the prompt that states that one of them 'must' be on the team, so logic dictates that we have to consider that possibility.
.
.
A and 3 of 5 others
B and 3 of 5 others
Neither and 4 of the other 5

GMAT assassins aren't born, they're made,
Rich

I have a doubt in the highlighted part. 6C4 * 2! = 15 * 2 = 30.
Can you please explain where the other 2 is from to get the total value of 60. Because the method followed gives the correct answer.

EMPOWERgmatRichC Bunuel chetan2u

Can anyone explain what is wrong with this approach:

a) No of ways to select consultants when 1 of the consultant is not included:

2C1 * 3C1 * 6C4 = This gives me 90 ways in which 1 consultant out of the 2 is not selected.

b) No of ways to select consultants when the second consultant is not included:

2C1 * 3C1 * 6C4 = This gives me 90 ways.

So total number of ways are 180

Can someone explain which case I am missing because of which the difference if 30 (180-150)?