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21 Aug 2019, 01:13
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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
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.
26 Sep 2019, 06:51
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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.
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.
26 Sep 2019, 12:58
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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
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
