A firm has 4 senior partners and 6 junior partners. How many different

statement in the brackets is very confusing. can anyone explain ?
as per me answer should be 96. we cannot consider the case third because all the member will be seniors and thus 4 groups wont be considered different.

Hi ashusharma2965,

The wording in parentheses is there to help us eliminate duplicate entries - for example, let's call the 4 senior partners A, B, C and D and the 6 junior partners 1, 2, 3, 4, 5 and 6.
The group "A/B/1" and the group "B/A/1" are the SAME group, since they have the same 3 members - thus, we are NOT supposed to count that group more than once.

Based on the wording in the prompt, there are 3 types of groups that "fit" what we are looking for.

1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups
2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups
3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups

4 + 36 + 60 = 100 different groups

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

Shouldn't we go for permutations rather than combination here ?

Hi dasoisheretorule,

Many questions that can be approached as a Permutation or Combination can ALSO be approached in the other way (it's just a matter of properly organizing your calculations). This prompt specifically refers to the number of "different GROUPS" (meaning that the 'order' of the members does NOT matter) - and that's a common clue that we're dealing with a Combination Formula question.

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

How can we use the FCP process to calculate this?

IMO, it can be something like this:
Total possible group - group with no senior partner
10*9*8 - 6*5*4 = 600, which is wrong.

But I don't understand why. Can someone please explain?

Hi soumyadeeppaul1,

In a Permutation, the order 'matters.' For example, if you had 10 people in a race, and you wanted to know the total number of possible winners for 1st/2nd/3rd place, then there would be (10)(9)(8) = 720 possible outcomes (assuming that there were no 'ties' during the race).

You're trying to use that same logic here, but the prompt asks us for 'groups' (NOT 'arrangements'), so you're ending up with too many 'duplicate entries' - and those duplicates are impacting your calculations.

Here's a simple example: if there were 3 people, then how many 'groups of 3' are there vs. how many different ways can you arrange the 3 people:

Let's refer to the people as A, B and C.
-Since there are only 3 people, there is only 1 possible 'group of 3' (A, B and C in whatever order you choose).
-With 3 people, there are (3)(2)(1) = 6 different arrangements (ABC, ACB, BAC, BCA, CAB and CBA).

There's a reason why your calculation is 6 TIMES larger than the correct answer: in both (10)(9)(8) and (6)(5)(4), each group is counted 6 TIMES (which each group should only be counted ONCE).

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

Okay, I think I get it.
So the first outcome (group of 3 people) is calculated by 3C3 = 3!/3!0! = 1
right?

Hi soumyadeeppaul1,

Yes - that's exactly how that particular calculation would turn out.

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

4 senior partners and 6 junior partners.

3 partners group with at least 1 senior partner: \(^4{C_1} * ^6{C_2} + ^4{C_2} * ^6{C_1} + ^4{C_3}\)

=> (4 * 15) + (6 * 6) + (4) = 60 + 36 + 4 = 100

Answer B

I love how you negated the at least 1 senior partner by using all junior partners.... Truly brilliant

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________