Math Is Fun Forum

coolwind

Member

Registered: 2005-10-30

Posts: 30

A combinations with repetition

(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)

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: A combinations with repetition

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

Posts: 3,791

Re: A combinations with repetition

for b, what the heck is n?

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"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..."

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: A combinations with repetition

im thinking b = n

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

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The Beginning Of All Things To End.
The End Of All Things To Come.

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: A combinations with repetition

for (a) can they be the same?

Hi,luca-deltodesco
  the ans is yes.

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

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: A combinations with repetition

im thinking b = n

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

What 's the different?:D

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A combinations with repetition

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.

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"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..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: A combinations with repetition

(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

Re: A combinations with repetition

(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?

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: A combinations with repetition

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.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: A combinations with repetition

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

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"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..."

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: A combinations with repetition

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

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

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: A combinations with repetition

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

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IPBLE:  Increasing Performance By Lowering Expectations.