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.

rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksss

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.

Thanks Rahul for above.
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.

N: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.

Thanks N:dure
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.

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.

anshumishra 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.

Thanks anshul for your post above
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.

kapur.arnav wrote:

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.

rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksss

Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :

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

anshumishra wrote:

kapur.arnav wrote:

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.

rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksss

Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :

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....

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!!

kapur.arnav wrote:

anshumishra wrote:

kapur.arnav wrote:

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.

rahul there is formula also for this if im not mistaken - n+r-1Cr-1... can u elaborate using that!!! thanksss

Lets calculate the number of ways in which 7 one $ bills can be distributed among 3 persons (where each can get 0 or more ) :

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....

Exactly. So, if you are interested in formula :


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