Combinatronics Law Firm Problem

[kouranjelika]

A certain lawfirm consists of 4 Senior partners and 6 Junior partners. How many dif groups of 3 can be formed where at least one senior partner on the team? (groups are considered different when at least one partner is different)

Here are my steps:
We have 10 TOTAL partners - 4 Senior 6 Junior
Let's first find the number of teams we can total and then subtract the teams which have no seniors on them (only juniors), since this does not fit our constraint.

All possibilities:
(10*9*8)/2*3 = 3*4*10 = 120
Here we are dividing by 2 AND 3 BECAUSE both of these partners may be found on such a team and therefore these elements are interchangeable.

ONLY Juniors:
(6*5*4)/3 = 40 >>>>and this is where I believe my logic is flawed, because (they don't provide the solution), but I believe they are also dividing here by 2.
I only divided by 3 because aren't the others not interchangeable since they are not even considered now and we only have 3 items being moved around? Please clarify here.

Then in theory I would go 120-40 = 80 dif teams. THIS IS INCORRECT. It is 100.

So then I tried a dif approach, very long and annoying but I want to understand this though roughly.
I tried this:

Team with 1 Senior: (4*9*8)/2 dividing by the two interchangeable juniors
OR
Team with 2 Seniors: (4*3*8)/2 dividing by the two interchangeable seniors now
OR
Team with 3 Seniors: (4*3*2)/3 dividing by the three interchangeable seniors here

144+48+8=200 Nice, round WRONG number.

Please explain this.

Also, Ron, do you remember doing more than one study hall on counting stuff? I only found the one from like 08 or whatever (one of the first ones out there).

Thanks!

[taliwarburg]

Your problem is in this step:

"ONLY Juniors:
(6*5*4)/3 = 40 >>>>and this is where I believe my logic is flawed, because (they don't provide the solution), but I believe they are also dividing here by 2.
I only divided by 3 because aren't the others not interchangeable since they are not even considered now and we only have 3 items being moved around? Please clarify here."

You should be dividing the number of "ONLY junior" teams by 3! not just by 3, since you do not want to double count some of the same teams. If you divide by 3! (like you did in the first step) you will get "20" which, when subtracted from 120, gives you 100.

Hope this helps!

Also, when you did the problem the second way you were adding the seniors into the pool from which you were drawing juniors: "Team with 1 Senior: (4*9*8)/2 dividing by the two interchangeable juniors." I think this should be (4*6*5/2).

[RonPurewal]

As the poster above points out, you should divide by 3!, not 3, in your first work-up.

Team with 1 Senior: (4*9*8)/2 dividing by the two interchangeable juniors
OR
Team with 2 Seniors: (4*3*8)/2 dividing by the two interchangeable seniors now
OR
Team with 3 Seniors: (4*3*2)/3 dividing by the three interchangeable seniors here

There are only six juniors, not nine or eight.

[RonPurewal]

By the way, please read the forum rules, and take care to follow them.

* The title of the post should consist of the first few words of the problem statement.

* Please post the entire problem, including all answer choices.
The answer choices are especially important in this problem. Once you determine that the total number of all possible three-person committees is 120, you can figure out the correct answer even if you don't know how to calculate it. (There are only two choices smaller than 120, and one of them is much too small to pass the proverbial smell test.)

[RonPurewal]

Last but not least, please search the forum before posting. Any GMAT Prep problem is virtually guaranteed to have an existing thread.

a-certain-law-firm-consists-of-4-senior-partners-t4110.html

If you have further questions about the problem, please post them there, not here. Thanks.