Number of objects distributed between four people

[gruba]

Homework Statement

Four people are dealing the total amount of money, which is $1000$ monetary units in terms of $100$ monetary units. Count the number of ways for this distribution if:
$1)$ Every person doesn't have to get any money
$2)$ Every person will get at least $100$ monetary units
$3)$ First person will get at least $500$ m.u. and other three people at least $100$ m.u.

Homework Equations

-Combinatorics

The Attempt at a Solution

The problem doesn't state what is the maximum amount of money that each person can get.
Assuming that, in $1)$ every person will get the same amount ($200$ m.u.), the total number of counts would be the coefficient with $x^{1000}$ in $(1+x+...+x^{200})^4$.

In $2)$, assuming that the maximum amount for every person is $200$ m.u, the total number of counts would be the coefficient with $x^{1000}$ in $(x^{100}+...+x^{200})^4$.

In $3)\Rightarrow x^{500}(x^{100}+...+x^{200})^3$

What do you think, how to solve this problem?

 

[andrewkirk]

You can convert (2) and (3) into the same problem as (1) with a different total to be distributed, by doing something non-probabilistic before you start the random part. What might that non-probabilistic step be?BTW, your formula $(1+x+...+x^{200})^4$ doesn't need to be have such a huge number of terms. Since everything is done in multiples of 100mu, divide the amounts by 100mu before you start.

 

[Ray Vickson]

Homework Statement

Four people are dealing the total amount of money, which is $1000$ monetary units in terms of $100$ monetary units. Count the number of ways for this distribution if:
$1)$ Every person doesn't have to get any money
$2)$ Every person will get at least $100$ monetary units
$3)$ First person will get at least $500$ m.u. and other three people at least $100$ m.u.

Homework Equations

-Combinatorics

The Attempt at a Solution

The problem doesn't state what is the maximum amount of money that each person can get.
Assuming that, in $1)$ every person will get the same amount ($200$ m.u.), the total number of counts would be the coefficient with $x^{1000}$ in $(1+x+...+x^{200})^4$.

In $2)$, assuming that the maximum amount for every person is $200$ m.u, the total number of counts would be the coefficient with $x^{1000}$ in $(x^{100}+...+x^{200})^4$.

In $3)\Rightarrow x^{500}(x^{100}+...+x^{200})^3$

What do you think, how to solve this problem?

I found the wording of your problem confusing: if you mean that each person can receive 0 or 100 or 200 or ... (i,e, multiples of 100) then you should say that a bit more clearly. If that is what you mean, it would be much easier to let 100 m.u be a new monetary unit. Now the amount each person can receive is an integer number of these new units and we have a total of 10 new units for distribution.

Also, to clarify: in (3), do you mean (in these new units) that persons 2--4 get at least 1 in total, or that persons 2--4 will get at least 1 each?

 

[haruspex]

1
Assuming that, in 1) every person will get the same amount (200 m.u.)

That would not be a good assumption. As I read it, the whole amount is to be divvied up, and there's no restriction on who gets what. Essentially, you have ten identical things to put in four (distinct) baskets.