BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Speeches

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2789
Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members
Location: Chennai, India
Thanked: 206 times
GMAT Score:640
by GmatKiss » Mon Aug 15, 2011 3:32 am
An oratorical society consists of six members, and at an upcoming meeting, the members will present a total of four speeches. If no member presents more than two of the four speeches, in how many different orders could the members give speeches?

(A)720
(B)1080
(C)1170
(D)1470
(E)1560
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Aug 15, 2011 8:12 am
GmatKiss wrote:An oratorical society consists of six members, and at an upcoming meeting, the members will present a total of four speeches. If no member presents more than two of the four speeches, in how many different orders could the members give speeches?

(A)720
(B)1080
(C)1170
(D)1470
(E)1560
4 different speakers:
Number of members who could give the first speech = 6.
Number of remaining members who could give the second speech = 5.
Number of remaining members who could give the third speech = 4.
Number of remaining members who could give the fourth speech = 3.
To combine the choices above, we multiply:
6*5*4*3 = 360.

3 different speakers:
One member must give a pair of speeches.
Number of pairs of speeches that can be formed from 4 choices = 4C2 = 6.
Number of members who could give the pair of speeches = 6.
Number of remaining members who could give one of the two remaining speeches = 5.
Number of remaining members who could give the one remaining speech = 4.
To combine the choices above, we multiply:
6*6*5*4 = 720.

2 different speakers:
The speeches can be divided as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Number of ways to divide the speeches = 3.
Number of members who could give the first pair of speeches = 6.
Number of remaining members who could give the remaining pair of speeches = 5.
To combine the choices above, we multiply:
3*6*5 = 90.

Total number of orderings = 360+720+90 = 1170.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 608
Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members
Thanked: 37 times

by saketk » Tue Aug 16, 2011 8:31 am
Similar to what Mitch has done here,

My approach--

1 speech per person-- this way we need 4 people out of 6. 4 people can be chosen in 6C4 ways.
and 4 speeches can be distributed in 4! ways. Total number of ways - 24* 15.= 360


second case - 2 speech per person
For this we need 2 people. This can be done in 6C2 ways.
Now let's say the two person selected are A and B
4 speeches can be distributed between A and B in 6 ways I.e in 4C2 ways
Total comes out to be 90


Third and last case
One person delivers 2 speech and rest 2 are divided between 2 people
Total number of people required 3
They can be selected in 6C3 ways
Now let's say they are A, B and C
We need to choose a person who will deliver 2 speech
This can be done in 3 ways, say A
The other 2 left are B and C
Now, A can choose 2 speech from 4 in 4C2 ways
The other 2 can chose from 2 speech in 2 ways
Therefore total comes out to be 20*3*6*2 = 720

do the total of 3 case and we will get our answer,
Which is C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 312
Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members
Location: New York City
Thanked: 130 times
GMAT Score:780

by gmatboost » Wed Aug 17, 2011 1:46 pm
The subtraction approach works well here.

Total possibilities 6*6*6*6 =1296

Ways that 1 person gives all 4 speeches: 6

Ways that 1 person gives 3 speeches:
Who gives the 3 speeches: 6 options
Which 3 speeches does he give? 4C3 = 4 options
Who gives the last speech: 5 options

6*4*5 = 120

1296 - 120 - 6 = 1170.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 94
Joined: Sun Aug 29, 2010 3:40 amFollowed by: 1 member
Thanked: 3 times

by leonswati » Thu Oct 06, 2011 7:08 am
Hi Mitch,

Thanks for the explanation.. But can you please explain how did u get this:

3 different speakers:
One member must give a pair of speeches.
Number of pairs of speeches that can be formed from 4 choices = 4C2 = 6.
Number of members who could give the pair of speeches = 6.
Number of remaining members who could give one of the two remaining speeches = 5.
Number of remaining members who could give the one remaining speech = 4.
To combine the choices above, we multiply:
6*6*5*4 = 720.

2 different speakers:
The speeches can be divided as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Number of ways to divide the speeches = 3.
Number of members who could give the first pair of speeches = 6.
Number of remaining members who could give the remaining pair of speeches = 5.
To combine the choices above, we multiply:
3*6*5 = 90.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members
Thanked: 114 times

by knight247 » Thu Oct 06, 2011 7:22 am
Edited.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members
Thanked: 114 times

by knight247 » Thu Oct 06, 2011 9:30 am
@Mitch
2 different speakers:
The speeches can be divided as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Number of ways to divide the speeches = 3.
Number of members who could give the first pair of speeches = 6.
Number of remaining members who could give the remaining pair of speeches = 5.
To combine the choices above, we multiply:
3*6*5 = 90.
This is the point I'm having trouble with. Assume the speakers are A B C D E F Now if I need two different speakers example AABB Then I can select two in 6C2= 15 Ways And the four speeches to be assigned to them can be done in 4C2=6 Ways. So possibilities=15*6=90

Is my reasoning correct?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2789
Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members
Location: Chennai, India
Thanked: 206 times
GMAT Score:640

by GmatKiss » Thu Oct 06, 2011 10:43 am
gmatboost wrote:The subtraction approach works well here.

Total possibilities 6*6*6*6 =1296

Ways that 1 person gives all 4 speeches: 6

Ways that 1 person gives 3 speeches:
Who gives the 3 speeches: 6 options
Which 3 speeches does he give? 4C3 = 4 options
Who gives the last speech: 5 options

6*4*5 = 120

1296 - 120 - 6 = 1170.
Hi am unable to follow the part in green!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Thu Oct 06, 2011 1:06 pm
knight247 wrote:@Mitch
2 different speakers:
The speeches can be divided as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Number of ways to divide the speeches = 3.
Number of members who could give the first pair of speeches = 6.
Number of remaining members who could give the remaining pair of speeches = 5.
To combine the choices above, we multiply:
3*6*5 = 90.
This is the point I'm having trouble with. Assume the speakers are A B C D E F Now if I need two different speakers example AABB Then I can select two in 6C2= 15 Ways And the four speeches to be assigned to them can be done in 4C2=6 Ways. So possibilities=15*6=90

Is my reasoning correct?
Looks good.

In my solution:
I counted the number of ways the speeches could be divided without regard to order (4C2/2! = 3).
I took order into account by counting the number of ways the speakers could be arranged (6*5 = 30).

In your solution:
You counted the number of ways the speakers could be selected without regard to order (6C2 = 15).
You took order into account by counting the number of ways the speeches could be assigned (4C2 * 2C2 = 6).

Either way is fine.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Mon Aug 29, 2011 6:43 pm
Thanked: 1 times

by razorback » Fri Oct 07, 2011 12:02 am
I understand how to arrange 6 people into 4 individual speeches (6! over 2! = 360).

I falter at the next step. If 3 people are speaking, (3 people speak once, 1 person gives a pair of speeches), then...??? What I did was run 6!/3!*3! to get 20 different groups of 3 speaks, then multiplied by 4!, then number of orders of speeches those 20 groups could speak in = 480.

Likewise I got 15 different groups of 2 speakers, multiplied by 4! for the order those speeches could go in, for 360. 360+360+480 = 1200. (frown face)

Also, just to clarify, in forum speak, 4C2 means 4!/2! correct?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1