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

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

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