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Manager
Joined: 31 Aug 2020
Posts: 56
Given Kudos: 39
Location: India
Concentration: Technology, General Management
GMAT 1: 710 Q48 V37
Re: A firm has 4 senior partners and 6 junior partners. How many different
[#permalink]
08 Oct 2020, 21:12
1
Kudos
Okay, I think I get it.
So the first outcome (group of 3 people) is calculated by 3C3 = 3!/3!0! = 1
right?
So the first outcome (group of 3 people) is calculated by 3C3 = 3!/3!0! = 1
right?
Re: A firm has 4 senior partners and 6 junior partners. How many different
[#permalink]
20 May 2017, 03:59
Bunuel wrote:
gdk800 wrote:
1) 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
OA is B.
Total # of different groups of 3 out of 10 people: \(C^3_{10}=120\);
# of groups with only junior partners (so with zero senior memeber): \(C^3_6=20\);
So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.
Answer: B.
Your explanation says that there can be all 3 senior members also as mentioned AT LEAST 1 senior so there can be 3 at most , FINE
But in the last line (2 groups are considered different if at least one group member is different), if we consider this , then it means we need atleast one member should be different in a group ,
So 4C1 * 6C2 (1 senior 2 junior) + 4C2 * 6C1 (2 senior and 1 junior) + 4C3 (Why this 4C3? As we need atleast one different member but in this 4C3 we will have all seniors and no junior and this will contradict the last line)
Please help?
Re: A firm has 4 senior partners and 6 junior partners. How many different
[#permalink]
09 Mar 2018, 11:33
I don't understand why this is not valid:
a) Pick one senior partner from 4
b) Pick 2 x further partners from 9 remaining senior & junior partners
4 (senior partner) * 9 (other partner) * 8 (other partner) = 288
Order of selection does not matter, so divide by # of ways to re-arrange 3 partners i.e. 3! = 6
288 / 6 = 48
But this is the wrong answer. Why?
a) Pick one senior partner from 4
b) Pick 2 x further partners from 9 remaining senior & junior partners
4 (senior partner) * 9 (other partner) * 8 (other partner) = 288
Order of selection does not matter, so divide by # of ways to re-arrange 3 partners i.e. 3! = 6
288 / 6 = 48
But this is the wrong answer. Why?