A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

The answer is 120 but I can't seem to come up with it! First I did 4*6*5, thinking the first slot has 4 options and the other two 6 and 5 but realized that we could have two senior partners or 3 as well. I can't come up with 120 though!

Any help would be much appreciated!

Based on my calculation given below I think answer is 100.

(1) First let's find out that out of 10 partners how many total group can be formed with 3 partners?

10!/(10-3)!*3! = 120

(2) Now out of 6 juniors how many teams are possible of only Junior partner.

6!/((6-3)!*3!) = 20

So 120-20 = 100 different teams can be formed where at least one partner is Senior.

Thanks!

this is the most efficient solution.

the key to this solution is to pick up on the fact that the COMPLEMENTARY EVENT - i.e., the desired event not happening - is easier to think about in this case. because the desired event - "at least one senior partner" - can happen in many different ways (1, 2, or 3 senior partners), it's easier to turn to the complementary event, which is exclusively 0 senior partners and 3 junior partners. once the # of ways is found for this complementary event, you can just subtract it from the total number of ways to find the desired number.

most combinatorics / probability problems formulated with the words "at least" will be best attacked by finding the probability (or combinations) for the complementary event.

--

if you're interested in a direct calculation, here's how to do it:
# of ways to get 1 senior partner and 2 junior partners: (4! / 1!3!) x (6! / 2!4!) = 4 x 15 = 60
# of ways to get 2 senior partners and 1 junior partner: (4! / 2!2!) x (6! / 1!5!) = 6 x 6 = 36
# of ways to get 3 senior partners and 0 junior partners: (4! / 3!1!) x (6! / 0!6!) = 4 x 1 = 4
add these up = 100

Really a sweet and simple way.. Thanks Ron..

Ron: I get the other two solutions fully.
HOWEVER, how come the following approach doesn't work

Since 1 member must be a senior partner
we are left with 9 individuals and the question becomes...how many different ways can we select 2 members from these 9 individuals

this can be represented as:
1 * (9!/(2!*7!)) = 36

how come we get a different answer this way? where am i going wrong with this approach?

Hi Sudaif,
I think your approach does not account for the number of ways the first senior partner can be selected ....

yeap - i considered that, but that gets you to 4 * 36 = 144
which is different from the instructor calc. above

what am i missing here??

nope. if you do 4 x 36, then you're going to repeat any combination in which there is more than one senior partner.

for instance: let's say that the senior partners are A, B, C, and D.
* yes, there are 36 combinations that include A.
* yes, there are 36 combinations that include B.
* the problem, though, is that you can't add these -- because then you're double-counting the combinations that include both of these people.

so, if you're going to go this route, you're going to have to figure out all the double- and triple-counted combinations, and subtract out the repeats.

from a student:



nope -- can't do this. as soon as you start talking about SP vs. JP, you have just created spots that are NOT interchangeable. when we say "interchangeable positions", they have to be exactly identical positions; any distinctions, such as SP vs. JP, nullifies this idea.
so, you've got to break this down into the separate cases with different numbers of SP's and JP's, and divide by the appropriate factorials.
i.e.
all 3 SP's --> all three spots are interchangeable. therefore, this yields (4x3x2)/3! = 4.
2 SP's and 1 JP --> only the two SP spots are interchangeable. therefore you have (4x3x6)/2! = 36.
1 SP and 2 JP's --> only the two JP spots are interchangeable. therefore you have (4x6x5)/2! = 60.

Hi

Sorry I still dont understand Why i cant do as below:
1S2J: 4*6*5
2S1J: 4*3*6
3S: 4*3*2


I somehow always end up reverting to the slot method

you're assuming order matters, i.e. that the first ones selected must be the senior partners..

_________________
Tim Sanders
Manhattan GMAT Instructor