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.

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:

Join us for a debate on how these elite MBA programs are different from each other, what are their strengths & weaknesses, and how to decide which one is better for you.

Join us for a debate between the two MBA experts on how these elite MBA programs are different from each other, what are their strengths & weaknesses, and how to decide which one is better for you.

Only 2000 test-takers score 760 or higher annually. These students not only secure admissions at top B-schools, but they also get an estimated $100M in scholarships. Find out more...

To score Q50 on GMAT Quant, you would need a strategic plan of action which takes your strengths and weaknesses into consideration. Attend this free workshop to attempt a supervised quant quiz and gain insights that can help you save 35+ hours...

In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 02:01

1

5

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

32% (01:26) correct 68% (01:55) wrong based on 31 sessions

HideShow timer Statistics

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 02:45

2

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

As P and Q have to have at least one ball, let us give one each to P and Q. Now we are left with 5 balls and they have to be distributed amongst P, Q, R and S.

So P+Q+R+S=5 Now the direct formula of putting three partitions in 5 balls makes it 5+3 or 8 out of which we have to select these 3 partitions. => 8C3=8*7*6/3!=56

D

Since 5 is not a big number, we can find each way of distribution as 1) 5,0,0,0 —4 way 2) 4,1,0,0 — 4!/2!=12 3) 3,2,0,0 — 12 ways 4) 3,1,1,0 — 12 ways 5) 2,2,1,0– 12 ways 6) 2,1,1,1– 4 ways Total 56 ways
_________________

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 02:57

1

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

Though i got it wrong, here is my solution P(1), Q(1), (R+S)[5]=1*1*6( because 5 balls can be distributed among R and S in (0,5),(1,4),(2,3)*(2) as they can either be in R or S=6 ways [P+Q](3) R+S(4)= one ball in P , two ball in Q or Reverse so 2 ways * for R and S ( 0,4),(1,3),(2,2) 5 ways as 2,2 is same for R or S=10 ways P+Q=4, R+S=3 Total 12 ways P+Q=(1,3),(2,2),(3,1) and R+S=(0,3),(1,2),(2,1),(3,0) P+Q=5, R+S=2 Total 12 ways P+Q=(1,4), (4,1),(2,3),(3,2) R+S= (0,2),(2,0),(1,1) P+Q=6, R+S=1 Total 10 ways P+Q= (1,5),(5,1),(3,3),(2,4),(4,2) R+S=(0,1),(1,0) P+Q=7 R+S=0 Total 6 ways P+Q= (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) Total=6+10+12+12+10+6 56 D:)

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 03:05

chetan2u wrote:

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

As P and Q have to have at least one ball, let us give one each to P and Q. Now we are left with 5 balls and they have to be distributed amongst P, Q, R and S.

So P+Q+R+S=5 Now the direct formula of putting three partitions in 5 balls makes it 5+3 or 8 out of which we have to select these 3 partitions. => 8C3=8*7*6/3!=56

D

Since 5 is not a big number, we can find each way of distribution as 1) 5,0,0,0 —4 way 2) 4,1,0,0 — 4!/2!=12 3) 3,2,0,0 — 12 ways 4) 3,1,1,0 — 12 ways 5) 2,2,1,0– 12 ways 6) 2,1,1,1– 4 ways Total 56 ways

Sir can you explain this highlighted part. Why we put 3 partition in 5. When there are 4 boxes.

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 03:13

1

yashikaaggarwal wrote:

chetan2u wrote:

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

As P and Q have to have at least one ball, let us give one each to P and Q. Now we are left with 5 balls and they have to be distributed amongst P, Q, R and S.

So P+Q+R+S=5 Now the direct formula of putting three partitions in 5 balls makes it 5+3 or 8 out of which we have to select these 3 partitions. => 8C3=8*7*6/3!=56

D

Since 5 is not a big number, we can find each way of distribution as 1) 5,0,0,0 —4 way 2) 4,1,0,0 — 4!/2!=12 3) 3,2,0,0 — 12 ways 4) 3,1,1,0 — 12 ways 5) 2,2,1,0– 12 ways 6) 2,1,1,1– 4 ways Total 56 ways

Sir can you explain this highlighted part. Why we put 3 partition in 5. When there are 4 boxes.

Posted from my mobile device

Hi

When we put 3 partitions, we get items in 4 places. oolololo.....lines show partition , so 2,1,1,1 loollooo.....so 0,2,0,3 And so on
_________________

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 03:24

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

given that out of 7 balls P,Q need to have atleast one ball each so we put 1 ball each in P,Q left with 5 balls which can be arranged in 4 boxes P,Q,R,S in (5+4-1) c ( 4-1) ways or say 8c3 ways OPTION D ; 56

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 03:26

Ruchirkalra wrote:

chetan2u wrote:

Now the direct formula of putting three partitions in 5 balls makes it 5+3 or 8 out of which we have to select these 3 partitions. => 8C3=8*7*6/3!=56

Can you please explain this a bit. Why are you doing 5+3?

There are 5 balls left and we can put them in any way. So ooooo, what we do is we take 3 partitions so that we can place those partitions between these 5 balls so that these 5 balls are distributed in 4 parts. ooooolll, so total we have 5+3 places and the 3 partitions can take any of the 8 locations. It can be lllooooo....0,0,0,5 loololoo....0,2,1,2 ollloooo....1,0,0,4 and so on

This is the direct method for such linear equations where you have sum of variables on one side and a numerical value on the other side
_________________

In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

Updated on: 31 May 2020, 05:05

2

2

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

\(P+Q+R+S = 7\)

\(P_{min} = 1\) \(Q_{min} = 1\)

Let's give one ball in each box P and Q, so now we have 5 more balls to distribute among 4 Boxes P, Q, R and S

i.e. New Equation, P+Q+R+S = 5

Now, we have find non-negative solution to this equation

For any equation a+b+c+d = n

where n = Total Balls to be distributed and r = 4 (Number of variables)

The number of Non-negative solutions \(= (n+r-1)C_{r-1}\)

i.e. Total Non-negative Integer solutions of P+Q+R+S = 5 will be \((5+4-1)C_{4-1} = 8C2 = 56\)

Answer: Option D

DERIVATION OF PARTITION RULE:

If you like it then please subscribe to my YouTube channel

_________________

Prepare with PERFECTION to claim Q≥50 and V≥40 !!! GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha) e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772 One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 04:55

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

P 4.......2........3........5........4

Q 1.......2.........2.......2........2

R 1.......2.........1.......0.........1

S 1.......1.........1.......0.........0

So \(\frac{4!}{3!} + \frac{4!}{3!}+\frac{4!}{2!} +\frac{4!}{2!} +\frac{4!}{1! }= 4+4+12+12+24 = 56\)

So D

I wonder if there is a shortcut solution to this question, cause it took me more than 4 min to figure out the answer

Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
[#permalink]

Show Tags

31 May 2020, 05:03

1

dave13 wrote:

MBADream786 wrote:

In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each? A.84 B.70 C.120 D.56 E.54

P 4.......2........3........5........4

Q 1.......2.........2.......2........2

R 1.......2.........1.......0.........1

S 1.......1.........1.......0.........0

So \(\frac{4!}{3!} + \frac{4!}{3!}+\frac{4!}{2!} +\frac{4!}{2!} +\frac{4!}{1! }= 4+4+12+12+24 = 56\)

So D

I wonder if there is a shortcut solution to this question, cause it took me more than 4 min to figure out the answer

This is a typical question based on partition rule. So if one needs to do it in faster way then only wa forward I see is the explanation I have mentioned above your post.

To understand Partition Rule, refer to the video attached.
_________________

Prepare with PERFECTION to claim Q≥50 and V≥40 !!! GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha) e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772 One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting