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

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

Ball Distribution

Expert replies
Source: — Problem Solving |

by Brent@GMATPrepNow » Wed Jul 01, 2015 8:50 am
nahid078 wrote:In how many ways 5 different balls can be distributed in three boxes?
This question is a little ambiguous.
I'm assuming that the 3 boxes are also different (call them boxes A, B and C)
Let's also call the balls V, W, X, Y and Z

So, take the task distributing balls and break it into stages.

Stage 1: place ball V in a box
There are 3 boxes, so, we can complete stage 1 in 3 ways

Stage 2: place ball W in a box
There are 3 boxes, so, we can complete stage 2 in 3 ways

Stage 3: place ball X in a box
There are 3 boxes, so, we can complete stage 3 in 3 ways

Stage 4: place ball Y in a box
There are 3 boxes, so, we can complete stage 4 in 3 ways

Stage 5: place ball Z in a box
There are 3 boxes, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 balls) in (3)(3)(3)(3)(3) ways ([spoiler]= 243 ways[/spoiler])

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by nahid078 » Wed Jul 01, 2015 9:13 am
But if it is said that, one box can contain only one ball. Then How many ways can we put three balls in three boxes?

5*4*3= 60

Would it be the answer?

Thanks :)
Join the discussion

by Brent@GMATPrepNow » Wed Jul 01, 2015 9:17 am
nahid078 wrote:But if it is said that, one box can contain only one ball. Then How many ways can we put three balls in three boxes?

5*4*3= 60

Would it be the answer?

Thanks :)
Sorry, I think you need to clarify that restriction. Are you saying that one box must have exactly one ball, and the other 2 boxes can have any number of balls?

If so, the answer is (3)(2)(2)(2)(2)

If you mean something else, please tell me.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by nahid078 » Wed Jul 01, 2015 10:07 am
No i said every box can contain only one ball. Thus last two balls will not be in the box.
Join the discussion

by Brent@GMATPrepNow » Wed Jul 01, 2015 10:13 am
nahid078 wrote:No i said every box can contain only one ball. Thus last two balls will not be in the box.
Then yes, the answer is (5)(4)(3)

Choose a ball to go in box A: 5 balls to choose from
Choose a ball to go in box B: 4 balls remaining to choose from
Choose a ball to go in box C: 3 balls remaining to choose from

Total = (5)(4)(3)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by nikhilgmat31 » Thu Jul 02, 2015 11:52 pm
The question asks -In how many ways 5 different balls can be distributed in three boxes?

means 3 boxes to have 5 balls combined

balls can

1,1,3
1,3,1
3,1,1
1,2,2
2,1,2
2,2,1

Please suggest.
Join the discussion

by Amrabdelnaby » Wed Nov 18, 2015 11:52 am
Hi Brent,

Thank you for the answer; however I have a little question.

Why did you distribute the boxes on the balls and not vice versa?

I thought about distributing the balls on the boxes

so i thought that we have three boxes, for the first one we can put one of 5 balls and for the second we can put one of four balls and for the third we can put one of three.

leaving me with the following calculation 5x4x3 = 60 ways.

Why is this wrong?
Brent@GMATPrepNow wrote:
nahid078 wrote:In how many ways 5 different balls can be distributed in three boxes?
This question is a little ambiguous.
I'm assuming that the 3 boxes are also different (call them boxes A, B and C)
Let's also call the balls V, W, X, Y and Z

So, take the task distributing balls and break it into stages.

Stage 1: place ball V in a box
There are 3 boxes, so, we can complete stage 1 in 3 ways

Stage 2: place ball W in a box
There are 3 boxes, so, we can complete stage 2 in 3 ways

Stage 3: place ball X in a box
There are 3 boxes, so, we can complete stage 3 in 3 ways

Stage 4: place ball Y in a box
There are 3 boxes, so, we can complete stage 4 in 3 ways

Stage 5: place ball Z in a box
There are 3 boxes, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 balls) in (3)(3)(3)(3)(3) ways ([spoiler]= 243 ways[/spoiler])

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Join the discussion

by Brent@GMATPrepNow » Wed Nov 18, 2015 1:03 pm
Amrabdelnaby wrote:Hi Brent,

Thank you for the answer; however I have a little question.

Why did you distribute the boxes on the balls and not vice versa?

I thought about distributing the balls on the boxes

so i thought that we have three boxes, for the first one we can put one of 5 balls and for the second we can put one of four balls and for the third we can put one of three.

leaving me with the following calculation 5x4x3 = 60 ways.

Why is this wrong?
Brent@GMATPrepNow wrote:
nahid078 wrote:In how many ways 5 different balls can be distributed in three boxes?
This question is a little ambiguous.
I'm assuming that the 3 boxes are also different (call them boxes A, B and C)
Let's also call the balls V, W, X, Y and Z

So, take the task distributing balls and break it into stages.

Stage 1: place ball V in a box
There are 3 boxes, so, we can complete stage 1 in 3 ways

Stage 2: place ball W in a box
There are 3 boxes, so, we can complete stage 2 in 3 ways

Stage 3: place ball X in a box
There are 3 boxes, so, we can complete stage 3 in 3 ways

Stage 4: place ball Y in a box
There are 3 boxes, so, we can complete stage 4 in 3 ways

Stage 5: place ball Z in a box
There are 3 boxes, so, we can complete stage 5 in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus distribute all 5 balls) in (3)(3)(3)(3)(3) ways ([spoiler]= 243 ways[/spoiler])

--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Which question are you referring to?
In my solution here, I am distributing the balls into the boxes.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion