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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

(a)Determine the number of integer solutions of

a1+a2+a3+a4=32

where a1,a2,a3>0, 0<a4<=25

(b)Mary has two dozen each of n different colored beads.

If she can select 20 beads(with repetition of colors allowed)

in 230,230 ways,what is the value of n?

Thanks

*Last edited by coolwind (2006-08-08 03:29:52)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

for (a) do they have to be distinct numbers, or can they be the same?

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**Ricky****Moderator**- Registered: 2005-12-04
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for b, what the heck is n?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

im thinking b = n

but also, are the beads put back in the mix, or kept seperate?

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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

luca-deltodesco wrote:

for (a) can they be the same?

Hi,luca-deltodesco

the ans is yes.

*Last edited by coolwind (2006-08-08 03:27:58)*

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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

luca-deltodesco wrote:

im thinking b = n

but also, are the beads put back in the mix, or kept seperate?

What 's the different?:D

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Something is wrong in b.

If she has 2 dozen of each (24), and she is only picking 20 beads total, then it doesn't matter how many of each she has. It could be infinite.

So she has n choices for the first bead, n choices for the second bead, n choices for the.... which is n*n*n....*n = n^20. So n^20 = 230230 which comes out to 1.85399, which must be wrong.

Unless I'm missing something.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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(a)

If a1!=a2!=a3!=a4, then: 3258

If not: 4475

If you want a1<=a2<=a3<=a4, you will get 242 different solutions.

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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

krassi_holmz wrote:

(a)

If a1!=a2!=a3!=a4, then: 3258

If not: 4475

If you want a1<=a2<=a3<=a4, you will get 242 different solutions.

Right,the ans is 4475.

How did you count?

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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

Ricky wrote:

Something is wrong in b.

If she has 2 dozen of each (24), and she is only picking 20 beads total, then it doesn't matter how many of each she has. It could be infinite.

So she has n choices for the first bead, n choices for the second bead, n choices for the.... which is n*n*n....*n = n^20. So n^20 = 230230 which comes out to 1.85399, which must be wrong.

Unless I'm missing something.

Hi,Ricky

the ans is n=7.

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**Ricky****Moderator**- Registered: 2005-12-04
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coolwind, are you sure she doesn't have 1 dozen of each?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**coolwind****Member**- Registered: 2005-10-30
- Posts: 30

Ricky wrote:

coolwind, are you sure she doesn't have 1 dozen of each?

Ricky,this problem is from my textbook(written by Ralph P.Grimaldi)

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I have a notebook in which the numberof solutions of such equations is given as a recursive formula and generating function.

IPBLE: Increasing Performance By Lowering Expectations.

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