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)

]]>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.

(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?

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

If not: 4475

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

]]>

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.

]]>im thinking b = n

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

What 's the different?:D

]]>for (a) can they be the same?

Hi,luca-deltodesco

the ans is yes.

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

]]>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