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

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A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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09 Dec 2010, 07:21
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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
[Reveal] Spoiler: OA
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09 Dec 2010, 08:38
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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

[Reveal] Spoiler:
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.

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20 Apr 2013, 10:22
What seems to work as well, though its tougher to come up with this solution:

(10*9*8) * 2/3 + (10*9*4)*1/3 = 600. Divided by all possible permutations (=3!) yields 100.

Explanation: For the first spot you have 10 possible canditates, for the 2nd 9. For the 3rd spot you need to differentiate if a senior partner has been picked for one of the first 2 spots or not. If yes, then you can pick one of the 8 remaining guys, so 10*9*8. If no senior partner has been picked yet, you need 10*9*4.
Now you need to weight the 2 cases with the corresponding probabilities: probability of no senior partner being picked in the first 2 draws = 6/10*5/9=1/3, so you weight 10*9*4 with 1/3. For the complementary case (senior partner was picked in the first 2 draws) you just take the complementary prob (1-1/3)= 2/3 and weight 10*9*8 with it.
Now you just need to divide the answer (600) by the number of different positions (=3!) and get 600/6=100

But I suggest you stick with the easy solution in the GMAT
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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20 Apr 2013, 17:57
The standard approach

1 senior partner
4 x 6C2 = 60

2 senior partners
4C2 x 6 = 36

3 senior partners
4C3 = 4

Total 100
However GDKs approach is a better way around
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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22 Apr 2013, 21:44
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

A general approach to solving Combination problems

Steps:

1. There are two larger groups, the senior partners and the junior partners.
2. The first larger group i.e., the senior partners is 4 in number. So let$$n1$$ be 4. The second larger group, ie, the junior partners is 6 in number. So $$n2$$ is 6.
3. The smaller group that is selected from the larger group of senior partners may be 1, 2, or 3 in number. So $$r1$$ is 1 or 2 or 3 . Correspondingly the other smaller group i.e.,$$r2$$ selected from the junior partners is 2 or 1 or 0 in number.
4. For each value of$$r1$$ and the corresponding$$r2$$, compute the number of combinations which are$$4C1 * 6C2$$, $$4C2 * 6C1$$ and$$4C3 * 6C0$$ being 60, 36 and 4 ways respectively.
5. The total number of combinations is therefore 60+36+4 = 100
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A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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28 Dec 2014, 07:49
You can also think about it as a code.

4S - 6J

three different ways to group them:

S J J. 4(6)5/2! because we have two junior members
or
S S J. 4(3)6/2! because we have two senior members
or
S S S 4(3)2/3! because we have three senior members.

add up the results and you get 100.
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Last edited by gmat6nplus1 on 28 Dec 2014, 09:38, edited 1 time in total.
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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28 Dec 2014, 09:07
Hi All,

This is a rare layered Combinatorics question. To answer it, you can either figure out the Total of ALL possibilities and subject the ones that "don't fit" or you can perform 3 separate calculations to account for the ones that "do fit." Here is how you can approach the latter option:

We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations:

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

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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** VP Joined: 08 Jun 2010 Posts: 1402 Followers: 3 Kudos [?]: 116 [0], given: 813 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 10 May 2015, 01:40 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 [Reveal] Spoiler: 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. Princeton book is wonderful for explanation of coounting and combination. read it to understand of the math legendary GMAT Club Legend Joined: 09 Sep 2013 Posts: 13509 Followers: 577 Kudos [?]: 163 [0], given: 0 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 24 Jun 2016, 11:34 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Director Joined: 18 Oct 2014 Posts: 924 Location: United States GMAT 1: 660 Q49 V31 GPA: 3.98 Followers: 93 Kudos [?]: 235 [0], given: 69 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 24 Jun 2016, 13:02 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 Ways to choose at least 1 Senior 1 S and 2 J 4C1*6C2= 60 2 S and 1 J 4C2 * 6C1= 36 3 S and 0 J 4C3 = 4 60+36+4= 100 B is the answer _________________ I welcome critical analysis of my post!! That will help me reach 700+ Intern Joined: 09 Oct 2016 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 0 A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 09 Oct 2016, 19:24 Hi, Can some one help me understand what is wrong with my logic here? A group should have 3 partners with at least 1 senior partner. Overall there are 4 senior partners & 6 junior partners. At least 1 senior partner = 4C1 Remaining 2 can be anyone from the remaining partners = (10-1)C2 4C1*9C2 = 144? Can some one help me understand what i am missing here? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 8328 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 382 Kudos [?]: 2471 [1] , given: 163 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 09 Oct 2016, 21:04 1 This post received KUDOS Expert's post Hi ashwinchivukula, Unfortunately, your math includes some '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. In your calculation, you state that the first person selected MUST be one of those 4 seniors (A/B/C/D) and the remaining two people can be any two of the remaining 9... The group "A/B/1" and "B/A/1" are the SAME group, but your calculation counts THAT group TWICE (depending on whether A or B was chosen first). In a Combination question, you can't allow duplicate entries. GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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18 Dec 2016, 04:29
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

[Reveal] Spoiler:
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.

Hi Bunuel

The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:

Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp

Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp

Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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18 Dec 2016, 22:09
Nightfury14 wrote:
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

[Reveal] Spoiler:
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.

Hi Bunuel

The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:

Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp

Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp

Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.

The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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18 Jan 2017, 16:21
Hello,

If one finds 48 by doing (4)(9)(8)/3!, what has he actually computed? Why is it incorrect to think that i can pick from 4 senior at first -obligatory-, then my choice is open to senior and junior (9 then 8). Why is this incorrect?
Re: A firm has 4 senior partners and 6 junior partners. How many   [#permalink] 18 Jan 2017, 16:21
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