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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Combo.

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 233
Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members
Location: New York
Thanked: 7 times
by yellowho » Tue Mar 01, 2011 9:54 pm
In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members
Thanked: 127 times

by Night reader » Tue Mar 01, 2011 10:29 pm
IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Tue Feb 08, 2011 8:55 pm
Location: Canada
Thanked: 7 times
GMAT Score:700

by BarryLi » Tue Mar 01, 2011 11:03 pm
[spoiler]Thank goodness the coins are identical.

For 0 coins, there is 1 combination: 0-0-0
For 1 coin, there are 3 combinations: 0-0-1, 0-1-0, 1-0-0
For 2 coins, there are 6 combinations: 0-0-2, 0-1-1, 0-2-0, 1-0-1, 1-1-0, 2-0-0

There may be a pattern of adding 1, 2, 3 to the total number of combinations, but I'll working out the case for 3 coins just to confirm this.

For 3 coins, there are 10 distributions: 0-0-3, 0-1-2, 0-2-1, 0-3-0, 1-0-2, 1-1-1, 1-2-0, 2-0-1, 2-1-0, 3-0-0

From the sequence 1, 3, 6, 10, the number of terms seems to be increasing by 2, 3, 4, etc.

0 coins: 1
1 coin: 3
2 coins: 6
3 coins: 10
4 coins: 15
5 coins: 21
6 coins: 28

Yes, the solution is 28.[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 586
Joined: Tue Jan 19, 2010 4:38 amFollowed by: 5 members
Thanked: 31 times
GMAT Score:730

by rohu27 » Tue Mar 01, 2011 11:21 pm
can we have a case here where a=b=c=0?
Night reader wrote:IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members
Thanked: 127 times

by Night reader » Tue Mar 01, 2011 11:24 pm
I thought about this :) there's no distribution then - just 0s for everyone
rohu27 wrote:can we have a case here where a=b=c=0?
Night reader wrote:IOM [spoiler]because some of three people (ABC) may receive zero coins, which are identical
1) A(6) B(0) C(0)
2) A(5) B(1) C(0)
3) A(5) B(0) C(1)
4) A(4) B(2) C(0)
5) A(4) B(0) C(2)
6) A(4) B(1) C(1)
7) A(3) B(3) C(0)
8) A(3) B(0) C(3)
9) A(3) B(2) C(1)
10) A(3) B(1) C(2)
11) A(2) B(4) C(0)
12) A(2) B(0) C(4)
13) A(2) B(3) C(1)
14) A(2) B(2) C(2)
15) A(2) B(1) C(3)
16) A(1) B(5) (0)
17) A(1) B(0) C(5)
18) A(1) B(1) C(4)
19) A(1) B(4) C(1)
20) A(1) B(3) C(2)
21) A(1) B(2) C(3)
22) A(0) B(6) C(0)
23) A(0) B(0) C(6)
24) A(0) B(5) C(1)
25) A(0) B(1) C(5)
26) A(0) B(4) C(2)
27) A(0) B(2) C(4)
28) A(0) B(3) C(3)

added some more and we are equal - total 28 ways



[/spoiler]
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members
Thanked: 147 times

by anshumishra » Wed Mar 02, 2011 3:54 am
yellowho wrote:In how many ways can 6 identical coins be distributed among Alex, Bea and Chad? Note: Some people may receive zero coins.


I don't have solution to this. Is this 28?
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items(≤ n) is : (n+r-1)C(r-1)]

So, here the number of ways = (6+3-1)C(3-1) = 8C2 = 7*8/2 = 28.
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1