dabaobao wrote:

gdk800 wrote:

A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)

A. 48

B. 100

C. 120

D. 288

E. 600

Bunuel VeritasKarishma I got a doubt regarding an alternate approach. Would really appreciate your help

Let Senior Partner = S, Junior Partner = J

Approach: Choose 1 S and then consider the other cases.

Ways of choosing 1 S: 4C1 = 4

Then we have 3 S left

Now the possible cases of picking 2 people from 3S and 6J are: 1) 1S + 1J 2) 0S + 2J 3) 2S + 0J

Total # for those 3 cases = 3C1 * 6C1 + 6C2 + 3C2 = 3*6 + 15 + 3 = 36

Total cases = 4 * 36 = 144

What am I doing wrong since the answer doesn't match the OA? Thanks!

You are double counting cases. Whenever you make partial selections from a group for another group, you double count. Here, you first selected an S for your team and then selected 1 S or 2S or 0S again from the group of S for your team.

So say the 4 S are Sa, Sb, Sc and Sd. You select Sa.

Now in your first case, you have to select another S. Say you select Sc. Then you select Ja from juniors.

Now think of another case. First you select Sc.

Now in your first case, you select Sa. Then you select Ja from juniors.

Both these teams are the same but you counted them as 2. That's your error.

_________________

Karishma

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