For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
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Source: Beat The GMAT — Problem Solving |
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksssRahul@gurome wrote:For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
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prachich1987
- Legendary Member
- Posts: 752
- Thanked: 20 times
- GMAT Score:700
Thanks Rahul for above.Rahul@gurome wrote:For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
But if we want to actually calculate the no. of ways for column A,will the answer be 4 ways or it would be more than that?
Plz advise.
Thanks Rahul for your help! I believe since he had 7 correct bills (as in no change) and 3 people to distribute it on, there'll always a remainder and someone will get more than the other.
@ prachich1987: I think there can be more than 4 (1,1,5) (5,1,1) (1,2,4) (1,4,2) (4,1,2)..etc but in the end both ways will yield equal no.s. I don't know how to calculate that though.
@ prachich1987: I think there can be more than 4 (1,1,5) (5,1,1) (1,2,4) (1,4,2) (4,1,2)..etc but in the end both ways will yield equal no.s. I don't know how to calculate that though.
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prachich1987
- Legendary Member
- Posts: 752
- Thanked: 20 times
- GMAT Score:700
Thanks N:dureN:Dure wrote:Thanks Rahul for your help! I believe since he had 7 correct bills (as in no change) and 3 people to distribute it on, there'll always a remainder and someone will get more than the other.
@ prachich1987: I think there can be more than 4 (1,1,5) (5,1,1) (1,2,4) (1,4,2) (4,1,2)..etc but in the end both ways will yield equal no.s. I don't know how to calculate that though.
There must be more than 4 ways depending on who gets which coins
For e.g set 1--(1,1,5)
it can be like,
Lucia-1 ,Gomez -1,& Demingo-5
or
Lucia-5 ,Gomez -1,& Demingo-1 & many other ways
I just want to know the way to calculate it.
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anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
Nice Question.
Actually we don't need to calculate anything.
Check out the Column B : There is no way that you can distribute 7 one dollar bill to three people without having one as at least 3 dollars. So, that means the requirements for Column A and Column B is same.
Hence, they are same.
Actually we don't need to calculate anything.
Check out the Column B : There is no way that you can distribute 7 one dollar bill to three people without having one as at least 3 dollars. So, that means the requirements for Column A and Column B is same.
Hence, they are same.
-
prachich1987
- Legendary Member
- Posts: 752
- Thanked: 20 times
- GMAT Score:700
Thanks anshul for your post aboveanshumishra wrote:Nice Question.
Actually we don't need to calculate anything.
Check out the Column B : There is no way that you can distribute 7 one dollar bill to three people without having one as at least 3 dollars. So, that means the requirements for Column A and Column B is same.
Hence, they are same.
But I am already clear about the problem posted & just wanted to know the total no. ways of calculating column A ways.
It has nothing to do with this problem.
Sorry for creating confusion.
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anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :kapur.arnav wrote:rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksssRahul@gurome wrote:For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
Direct formula : n+r-1Cr-1 = 7+3-1C3-1 = 9C2
Other way :
We have these seven dollars and want to distribute to A, B and C : |||||||
So we need two separators to separate the sum (I am using *) : One possible distribution can be shown as :
|| * || * ||| (means A gets 2 $, b gets 3$, and C gets 4$)
You can think of it as a collection of 9 symbols ( with 2 * and 7 "|)
So, the problem reduces to selecting positions of two *, out of 9 symbols = 9C2 (which is same as the formula).
Please note, in this case we have not put the restriction that all the people get at least 1$.
Give it a try!
You can use the same technique with the restrictions that all get at least 1$.
I have left for you to try and let us know.
Thanks
That should be 4+3-1C3-1... as we need to account for atleast 1 to each... we will take n as = 4 and not as 7....anshumishra wrote:Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :kapur.arnav wrote:rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksssRahul@gurome wrote:For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
Direct formula : n+r-1Cr-1 = 7+3-1C3-1 = 9C2
Other way :
We have these seven dollars and want to distribute to A, B and C : |||||||
So we need two separators to separate the sum (I am using *) : One possible distribution can be shown as :
|| * || * ||| (means A gets 2 $, b gets 3$, and C gets 4$)
You can think of it as a collection of 9 symbols ( with 2 * and 7 "|)
So, the problem reduces to selecting positions of two *, out of 9 symbols = 9C2 (which is same as the formula).
Please note, in this case we have not put the restriction that all the people get at least 1$.
Give it a try!
You can use the same technique with the restrictions that all get at least 1$.
I have left for you to try and let us know.
Thanks
Hahhahah!!
When did BTG becanme beattheGRE??/ Eric r u starting a new website?? U never said us so!!
@Rahul Dada,
I am just immensely surprised by ur quant instincts...superb!!
When did BTG becanme beattheGRE??/ Eric r u starting a new website?? U never said us so!!
@Rahul Dada,
I am just immensely surprised by ur quant instincts...superb!!
-
anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
Exactly. So, if you are interested in formula :kapur.arnav wrote:That should be 4+3-1C3-1... as we need to account for atleast 1 to each... we will take n as = 4 and not as 7....anshumishra wrote:Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :kapur.arnav wrote:rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksssRahul@gurome wrote:For column A, possible distributions are (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
For calculating column B, let us assume that no person gets $3.
So the maximum a person can get is $2.
Even if all 3 get $2,the sum will be 2+2+2 = 6 < 7.
So one person will have to get at least $3.
Hence ultimately, for column B we will get the same distributions : (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 3, 2).
So both quantities are equal in both the columns.
Direct formula : n+r-1Cr-1 = 7+3-1C3-1 = 9C2
Other way :
We have these seven dollars and want to distribute to A, B and C : |||||||
So we need two separators to separate the sum (I am using *) : One possible distribution can be shown as :
|| * || * ||| (means A gets 2 $, b gets 3$, and C gets 4$)
You can think of it as a collection of 9 symbols ( with 2 * and 7 "|)
So, the problem reduces to selecting positions of two *, out of 9 symbols = 9C2 (which is same as the formula).
Please note, in this case we have not put the restriction that all the people get at least 1$.
Give it a try!
You can use the same technique with the restrictions that all get at least 1$.
I have left for you to try and let us know.
Thanks
The number of ways of dividing n identical items among r persons or objects,with no restrictions : n+r-1Cr-1
The number of ways of dividing n identical items among r persons or objects, each one of whom receives at least 1 item :
n-1Cr-1, So here it is 7-1C3-1 = 6C2
How to derive it using the illustration I provided :
Just give one "|" each to A, B and C, before considering any distribution.
So, now the problem reduces to distributing 4 "|" and 2 "*" among A,B and C, with no restrictions.
Hence , as above it would be : 6C2.
Thanks
