Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

09 Dec 2010, 07:21

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

71% (02:16) correct
29% (01:56) wrong based on 300 sessions

HideShow timer Statistics

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)

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

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.

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

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

Show Tags

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
_________________

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

Show Tags

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.
_________________

learn the rules of the game, then play better than anyone else.

Last edited by gmat6nplus1 on 28 Dec 2014, 09:38, edited 1 time in total.

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

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

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
_________________

visit my facebook to help me. https://www.facebook.com/thang.thang.73307634

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.
_________________

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+

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.

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

Show Tags

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

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.

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.
_________________

Good luck ========================================================================================= "If a street performer makes you stop walking, you owe him a buck" "If this post helps you on your GMAT journey, drop a +1 Kudo "

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

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.

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

Show Tags

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?

gmatclubot

Re: A firm has 4 senior partners and 6 junior partners. How many
[#permalink]
18 Jan 2017, 16:21

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...