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.

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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