A certain law firm consists of 4 senior partners

[sash23]

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!

[cdihenia]

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.

[sash23]

Thanks!

[RonPurewal]

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.

this is the most efficient solution.

the key to this solution is to pick up on the fact that the "complementary event" -- in normal-person talk, that the desired event will NOT happen -- is, in this case, easier to think about.

see, the desired event, "at least one senior partner", can happen in many different ways (1, 2, or 3 senior partners).
on the other hand, there's only one way for this NOT to happen: namely, 0 senior partners and 3 junior partners. once you've found the # of ways in which that can happen, you can just subtract from the total number of possibilities to find the desired number.

incidentally, this approach will quite often-- in fact, i'll just come out and say "almost always"-- be the most efficient way to tackle problems containing the words "at least".

--

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

[maaz_gmat]

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

[sudaif]

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?

[debmalya.dutta]

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

[sudaif]

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??

[RonPurewal]

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.

[RonPurewal]

from a student:

This question has been solved but I have a specific question that has NOT been addressed. My exam is next week - will appreciate expedited response.

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

When I looked at this question, I wanted to apply the SLOT method. And this is how I thought about it
3 slots: we must have one senior partner, and the other 2 slots can be filled by any body
4*9*8=288
Now, since order does NOT matter -- we're picking "groups" of 3 here...we must divide by the factorial of the number of inter-changeable items.
There are 3 inter-changeable items.
Let senior partner = SP and junior partner = JP
SP, JP, JP
JP, SP, JP
JP, JP, SP
*are all the same
Thus, we divide 288/3! = 48
Note: my interpretation of a group is that it doesn't matter how the 3 items are arranged within that group. And it doesn't matter what those 3 items are.
The OA is 100.

HOW and WHY is my approach incorrect?

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.

[nehajadoo]

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

[tim]

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

[dips]

Why does the order matter between senior(s) and junior(s). That is why can't we calculate the number of groups as (4 * 9 * 8)/ (3 * 2 * 1).
Why can't all members be at same level, such that there is no order between the members of the group. Nothing in the question explicitly states that order between seniors and junior matters within the group.

[RonPurewal]

Why does the order matter between senior(s) and junior(s). That is why can't we calculate the number of groups as (4 * 9 * 8)/ (3 * 2 * 1).
Why can't all members be at same level, such that there is no order between the members of the group. Nothing in the question explicitly states that order between seniors and junior matters within the group.

Before this question can be answered, you need an understanding of what you're actually doing when you divide by factorials.
When you divide by a factorial in these calculations, you're accounting for repetitions of the same group of people, in different orders. We don't want those to be counted separately, so we divide by the number of repetitions.
For instance, if we were making a totally unencumbered choice of 3 people from 10, that would be (10*9*8)/3!. Here, the 3! represents the number of different times every group would appear if you made a complete list of everyone in every possible order. (For instance, if you had people A, B, C, then you'd see ABC, ACB, BCA, BAC, CAB, and CBA.)

What ruins this approach, here, is that your first choice is NOT free and unencumbered"”you're trying to restrict your first choice to seniors.
This is problematic, because different groups are now going to show up different numbers of times.

Say A, B, C are all seniors. Then any of them could be your first pick, so this group will show up in 6 different ways (ABC, ACB, BCA, BAC, CAB, and CBA). So, dividing by 3! would make sense for groups of three seniors.

Now, say A is a senior and Y and Z are juniors.
Because you're only looking at groups in which a senior is picked first, you will not see this group 6 times. You'll only see it twice: AYZ and AZY.

You can see the problem here: In this approach, dividing by 6! is inappropriate for any group containing any junior(s).

[Kholoud]

the problem stated that 2 groups are considered different if at least one group member is different.

why didn't we use permutation instead of combination?