Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 24 Dec 2012
Posts: 25
Concentration: Entrepreneurship, Social Entrepreneurship
WE: Design (Computer Software)

In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
28 Apr 2014, 00:59
Question Stats:
26% (01:25) correct 74% (00:56) wrong based on 433 sessions
HideShow timer Statistics
In how many ways can 5 different marbles be distributed in 4 identical pockets? A. 24 B. 51 C. 120 D. 625 E. 1024
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 09 Mar 2014
Posts: 53
Location: India
Concentration: General Management, Operations
GPA: 3.2
WE: Engineering (Energy and Utilities)

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
Updated on: 14 Jun 2014, 05:49
tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 Hi According to the problem we have to distribute 5 DIFFERENT marbles (say M1, M2, M3, M4, M5) into 4 IDENTICAL pockets. Since, the pockets are identical we need not worry about distributing the marbles to different pockets...we have to simply divide 5 marble into four or less groups. Therefore we can have following condition. 1. (5,0, 0, 0) > 1 possibility (Becase the pockets are identical, there can be only one case. (5,0,0,0) and (0,5,0,0) are identical. 2. (4, 1, 0, 0) >5C4 = 5 possibilities 3. (3, 2, 0, 0) > 5C3 = 10 possiblities 4. (3, 1, 1, 0) > 5C3 = 10 possiblities. There will be only 5C3 cases and not 5C3*2C1 because pockets are identical. For example. ((M1M2M3), M4, M5, 0) is same as ((M1M2M3), M4, 0, M5). Therefore just by grouping 3 out of 5 marbles we will get all the possibilities. 5. (2, 2, 1, 0) > 5C2*3C2/2 = 15 possibilities 6, (2, 1, 1, 1) >5C2 = 10 possibilities Total = 1+5+10+10+15+10 = 51 (Answer) Note: we cannot form (1, 1, 1, 1, 1) because there are only 4 pockets.
Originally posted by utkmittal1 on 28 Apr 2014, 10:58.
Last edited by utkmittal1 on 14 Jun 2014, 05:49, edited 1 time in total.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
28 Apr 2014, 21:44
tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 Here are two posts you will find useful: http://www.veritasprep.com/blog/2011/12 ... 93part1/http://www.veritasprep.com/blog/2011/12 ... spartii/They discuss distributing 5 distinct things in 4 distinct groups 5 identical things in 4 distinct groups 5 distinct things in 4 identical groups (the case in point) 5 identical things in 4 identical groups
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
28 Apr 2014, 21:47
tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets. 5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how. {5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits. {4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits. {3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits. {3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits. {2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits. {2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits. Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



SVP
Joined: 06 Sep 2013
Posts: 1885
Concentration: Finance

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
30 May 2014, 14:46
VeritasPrepKarishma wrote: tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets. 5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how. {5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits. {4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits. {3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits. {3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits. {2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits. {2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits. Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways Hi Karishma, I read your posts but still it isn't clear to me how one needs to count each event instead of just applying a combinatorics formula such as 'n+r1 C r1'? Bunuel, you are also invited to this discussion since I basically stole this formula from you Many thanks Cheers J



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 893
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
01 Jun 2014, 11:18
I don't think such question will come in exam; it takes too much analysis case by case and definitely we have no short cut to solve this.
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
01 Jun 2014, 20:41
jlgdr wrote: VeritasPrepKarishma wrote: tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets. 5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how. {5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits. {4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits. {3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits. {3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits. {2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits. {2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits. Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways Hi Karishma, I read your posts but still it isn't clear to me how one needs to count each event instead of just applying a combinatorics formula such as 'n+r1 C r1'? Bunuel, you are also invited to this discussion since I basically stole this formula from you Many thanks Cheers J (n+r1)C(r1) gives the number of ways of distributing n identical things in r distinct groups  this questions distributes n distinct marbles in r identical groups. For identical groups, you need to count.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 04 May 2014
Posts: 14
Location: India
GMAT 1: 750 Q51 V40 GMAT 2: 710 Q50 V36
GPA: 3.8

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
14 Jun 2014, 03:53
utkmittal1 wrote: tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 Hi According to the problem we have to distribute 5 DIFFERENT marbles (say M1, M2, M3, M4, M5) into 4 IDENTICAL pockets. Since, the pockets are identical we need not worry about distributing the marbles to different pockets...we have to simply divide 5 marble into four or less groups. Therefore we can have following condition. 1. (5,0, 0, 0) > 1 possibility (Becase the pockets are identical, there can be only one case. (5,0,0,0) and (0,5,0,0) are identical. 2. (4, 1, 0, 0) >5C4 = 5 possibilities 3. (3, 2, 0, 0) > 5C3 = 10 possiblities 4. (3, 1, 1, 0) > 5C3 = 10 possiblities. There will be only 5C3 cases and not 5C3*2C1 because pockets are identical. For example. ((M1M2M3), M4, M5, 0) is same as ((M1M2M3), M4, 0, M5). Therefore just by grouping 3 out of 5 marbles we will get all the possibilities. 5. (2, 2, 1, 0) > 5C2*3C2 = 15 possibilities 6, (2, 1, 1, 1) >5C2 = 10 possibilities Total = 1+5+10+10+15+10 = 51 (Answer) Note: we cannot form (1, 1, 1, 1, 1) because there are only 4 pockets. I like your detailed explanation. However, calculation for case no 5 seems incorrect. It would give 30. We need to divide by 2 to avoid duplicating cases like M1 M2 , M3 M4 , M5 , 0 and M3 M4 , M1 M2, M5 , 0 since the groups are identical. Thanks Arpit
_________________
Risk Everything!



Manager
Joined: 09 Mar 2014
Posts: 53
Location: India
Concentration: General Management, Operations
GPA: 3.2
WE: Engineering (Energy and Utilities)

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
14 Jun 2014, 05:48
ValKilmer wrote: I like your detailed explanation. However, calculation for case no 5 seems incorrect. It would give 30. We need to divide by 2 to avoid duplicating cases like M1 M2 , M3 M4 , M5 , 0 and M3 M4 , M1 M2, M5 , 0 since the groups are identical.
Thanks Arpit
Hi ValKilmiler Thanks for correcting me. Don't know how I missed to divide it by 2. Updated my post.



Manager
Joined: 10 Mar 2014
Posts: 212

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
06 Nov 2014, 02:33
VeritasPrepKarishma wrote: tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
a) 24
b) 51
c) 120
d) 625
e) 1024 The solution to your problem is given below. It distributes 5 distinct fruits into 4 identical baskets. 5 fruits can be split into 4 groups in the following ways: {5, 0, 0, 0}, {4, 1, 0, 0}, {3, 2, 0, 0}, {3, 1, 1, 0}, {2, 2, 1, 0}, {2, 1, 1, 1} Does it concern us that the baskets are identical? It does. Let’s see how. {5, 0, 0, 0} means that one basket has all 5 fruits and the rest of the 3 baskets are empty. It doesn’t matter which basket has the fruits because all the baskets are identical. So, this gives us 1 way of distributing the fruits. {4, 1, 0, 0} means that one basket has 4 fruits, another has the leftover 1 fruit and the other 2 baskets have no fruit. The lone fruit can be chosen in 5 ways. The rest of the 4 fruits will be together in another basket and 2 baskets will be empty. This gives us 5 different ways of distributing the fruits. {3, 2, 0, 0} means that one basket has 3 fruits, another has the leftover 2 fruits and the other 2 baskets have no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be together in another basket in one way and 2 baskets will be empty. This gives us 10 different ways of distributing the fruits. {3, 1, 1, 0} means that one basket has 3 fruits, another two have a fruit each and the leftover basket has no fruit. The 3 fruits can be chosen in 5*4*3/3! = 10 ways (= 5C3). The rest of the 2 fruits will be in two baskets in one way (since the baskets are all identical) and the last basket will be empty. This gives us 10 different ways of distributing the fruits. {2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits. {2, 1, 1, 1} means that one basket has 2 fruits and the rest of the 3 baskets have a fruit each. We can select 2 fruits out of 5 in 5*4/2! = 10 ways (= 5C2). This gives us 10 different ways of distributing the fruits. Total number of different ways of distributing the fruits = 1 + 5 + 10 + 10 + 15 + 10 = 51 ways Hi Karishma, Could you please provide your comments on following scenario. I am getting 30 in this instead of 15 {2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.Thanks



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
11 Nov 2014, 01:54
PathFinder007 wrote: Hi Karishma,
Could you please provide your comments on following scenario. I am getting 30 in this instead of 15
{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.
Thanks
As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to unarrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15 If you present your calculation, I can tell you what you are missing.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 22 Jul 2015
Posts: 5

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
04 Sep 2015, 15:03
VeritasPrepKarishma wrote: PathFinder007 wrote: Hi Karishma,
Could you please provide your comments on following scenario. I am getting 30 in this instead of 15
{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.
Thanks
As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to unarrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15 If you present your calculation, I can tell you what you are missing. I am getting 30 as well, here is my reasoning: [2 2 1 0]  I can select the first 2 out of 5 in 5C2=10 ways  for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30 What is wrong with this approach?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
05 Sep 2015, 04:06
gmatminator wrote: VeritasPrepKarishma wrote: PathFinder007 wrote: Hi Karishma,
Could you please provide your comments on following scenario. I am getting 30 in this instead of 15
{2, 2, 1, 0} means that 2 baskets have 2 fruits each, one basket has one fruit and the last basket is empty. We can select 1 fruit out of 5 in 5 ways. Now we are left with 4 fruits which have to be split into 2 groups of 2 each. This can be done in 4!/2!*2!*2! = 3 ways (We have already discussed this concept in the post on Groups. Check out the initial theory and question no. 2) This gives us 5*3 = 15 different ways of distributing the fruits.
Thanks
As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to unarrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15 If you present your calculation, I can tell you what you are missing. I am getting 30 as well, here is my reasoning: [2 2 1 0]  I can select the first 2 out of 5 in 5C2=10 ways  for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30 What is wrong with this approach? This is fine but you missed the last step. You have selected "first two" and "second two" i.e. you have arranged them in first and second. You need to unarrange them. So you have to divide by 2. You get 30/2 = 15
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 12 Oct 2014
Posts: 55
Location: India
Concentration: Finance, General Management
WE: Analyst (Investment Banking)

In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
07 Sep 2015, 21:39
What if, I just reverse the problem ? In how many ways can 4 different marbles be distributed in 5 identical pockets? Now, in this scenario do we have to select pockets first, which is 5C4 and then put marbles in 4! ? Any thoughts. Regards, Gaurav



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
07 Sep 2015, 23:29
GauravSolanky wrote: What if, I just reverse the problem ? In how many ways can 4 different marbles be distributed in 5 identical pockets? Now, in this scenario do we have to select pockets first, which is 5C4 and then put marbles in 4! ? Any thoughts. Regards, Gaurav If the pockets are identical, how would you select 4 out of the 5? You can pick 4 of the 5 identical pockets in just 1 way. You have to decide how to distribute marbles and then find the number of ways you can do that. (4, 0, 0, 0, 0)  All 4 marbles in 1 pocket in 1 way. All pockets are identical. (3, 1, 0, 0, 0)  Select 1 of the 4 marbles in 4C1 ways = 4 ways and put it separately in one pocket. Put other 3 marbles in any other pocket in 1 way. So total 4 ways. (2, 2, 0, 0, 0)  We have to split 4 marbles in 2 groups of 2 marbles each. This can be done in 4!/(2!*2!*2!) = 3 ways (2, 1, 1, 0, 0)  Select 2 of the 4 marbles in 4C2 = 6 ways and put them in one pocket. Put the other 2 marbles in other 2 pockets in 1 way. Total number of ways = 6 (1, 1, 1, 1, 0)  Put the 4 marbles individually in any 4 pockets in 1 way. This gives us a total of 1 + 4 + 3 + 6 + 1 = 15 ways Check out this post for a similar question (Question 3): http://www.veritasprep.com/blog/2011/12 ... spartii/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 13 Mar 2011
Posts: 21

In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
22 Mar 2016, 04:12
VeritasPrepKarishma wrote: gmatminator wrote: VeritasPrepKarishma wrote:
As explained above, you select one fruit in 5 ways. The rest of the 4 fruits are split into two groups (not distinct groups but we will worry about that later). Select 2 fruits out of 4 in 4C2 = 6 ways for group 1. Rest of the two fruits are group 2. But we need to unarrange the groups so splitting 4 fruits in 2 groups of 2 each can be done in 6/2 = 3 ways. Hence total number of ways = 5*3 = 15
If you present your calculation, I can tell you what you are missing.
I am getting 30 as well, here is my reasoning: [2 2 1 0]  I can select the first 2 out of 5 in 5C2=10 ways  for the second 2 I can select 2 out of 3 3C2= 3 way Multiplying 10*3=30 What is wrong with this approach? This is fine but you missed the last step. You have selected "first two" and "second two" i.e. you have arranged them in first and second. You need to unarrange them. So you have to divide by 2. You get 30/2 = 15 If we need to unarrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's unarrange them => 10/2! = 5 Please, tell me what point did I miss.
_________________
I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times!  Bruce Lee
Please, press the +1 KUDOS button , if you find this post helpful



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
22 Mar 2016, 08:43
leeto wrote: If we need to unarrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's unarrange them => 10/2! = 5
Please, tell me what point did I miss. Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover. In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 13 Mar 2011
Posts: 21

In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
Updated on: 29 Mar 2016, 00:16
VeritasPrepKarishma wrote: leeto wrote: If we need to unarrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's unarrange them => 10/2! = 5
Please, tell me what point did I miss. Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover. In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here. "In case we are splitting 4 in 2 groups of 2, we get two identical cases." To make thinks easier to member and eventually digest, could I assume that this is some kind of the "Mississippi Problem" logic ? So, for example, let say the same problem only  we have 6 fruits and 4 baskets and in case {3,3, 0, 0 }  I will divide all out expression on (2!) because we get two identical cases => I mean the whole expression probably will be (6C3*3C3) / (2!) = 10 And, in the same problem with 10 fruits and 4 baskets, in the case {3,3,2,2}  We'll divide all expression on (2!)* (2!) ( because two identical cases for 3 fruits and two identical cases for 2 fruits  So it will be look like (10C3*7C3*4C2*2C2) / (2! * 2! ) = 60 * 35 * 3 = 60*105 = 6300 Do I correctly understand or I miss something again ? P.S. Excuses me for my meticulousness and many thanks for your helps.
_________________
I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times!  Bruce Lee
Please, press the +1 KUDOS button , if you find this post helpful
Originally posted by leeto on 23 Mar 2016, 08:50.
Last edited by leeto on 29 Mar 2016, 00:16, edited 1 time in total.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India

Re: In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
28 Mar 2016, 21:36
leeto wrote: VeritasPrepKarishma wrote: leeto wrote: If we need to unarrange them in the case [2 2 1 0], why we didn't perform the same stem in other cases? For example, if I follow the logic, in the case {3, 2, 0, 0} => 5C3*2C2 = 10 => now let's unarrange them => 10/2! = 5
Please, tell me what point did I miss. Putting 3 in a basket is different from putting 2 in a basket so we select 3 out of 5 and put those in one basket. The rest of the 2 we put in another basket and that's it. You are selecting 3 the first time and have 2 leftover. In case we are splitting 4 in 2 groups of 2, we get two identical cases. You are selecting 2 the first time and have 2 leftover. So, putting A and B in a basket and C and D in another basket is the same as putting C and D in one basket and A and B in another basket. Hence we divide by 2! here. "In case we are splitting 4 in 2 groups of 2, we get two identical cases." To make thinks easier to member and eventually digest, could I assume that this is some kind of the "Mississippi Problem" logic ? So, for example, let say the same problem only  we have 6 fruits and 4 baskets and in case {3,3, 0, 0 }  I will divide all out expression on (2!) because we get two identical cases => I mean the whole expression probably will be (6C3*3C3) / (2!) = 10 And, in the same problem with 10 fruits and 4 baskets, in the case {3,3,2,2}  We'll divide all expression on (2!)* (2!) ( because two identical cases for 3 fruits and two identical cases for 2 fruits  So it will be look like (10C3*7C3*4C2*2C2) / (2! * 2! ) = 60 * 35 * 3 = 60*105 = 6300 Do I correctly understand or I miss something again ? P.S. Excuses me for my meticulousness and many thanks for your helps. Yes, you are right. You have understood the concept.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Joined: 17 Dec 2012
Posts: 635
Location: India

In how many ways can 5 different marbles be distributed in 4 [#permalink]
Show Tags
11 Jul 2016, 18:09
tusharGupta1 wrote: In how many ways can 5 different marbles be distributed in 4 identical pockets?
A. 24 B. 51 C. 120 D. 625 E. 1024 The marbles only need to be distinguished and so the problem boils down to selection of marbles grouped in different ways. There are 6 different types of grouping possible. Case 1: 5 marbles together. The 5 marbles can be selected in 1 way. Case 2: 4 marbles together and 1 marble separately. The marbles can be selected in 5C4 ways= 5 ways Case 3: 3 marbles together and 2 marbles together. The marbles can be selected in 5C3 ways=10 ways Case 4: 3 marbles together and two 1 marble each. The marbles can be selected in 5C3 ways=10 ways Case 5: two set of 2 marbles and 1 marble. The marbles can be selected in 5C2*3C2/2! ways = 15 ways case 6: 2 marbles together and three 1 marble. The marbles can be selected in 5C2 ways= 10 ways Total 51 In the case of 5, to clarify see that (1,2) (3,4) 5 and (3,4) (1,2) and 5 are the same and so on. This is because we are selecting 2 marbles twice i.e., first 5C2 and then 3C2.. So we divide by 2!.
_________________
Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com/bestonlinegrepreparation.php
Improve Intuition and Your Score Systematic Approaches




In how many ways can 5 different marbles be distributed in 4
[#permalink]
11 Jul 2016, 18:09



Go to page
1 2
Next
[ 21 posts ]



