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Columba has two dozen each of n different coloured beads. If she can select 20 beads (with repetition of colors allowed) in 230,230 ways, what is the value of n?

I'm trying to figure it out, but I'm having some trouble. I know that the question is asking how many colors are there, but I'm not sure how to go about getting there. Is there an applicable formula? And if so, what's it called?

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This is stars-and-bars problem in reverse, so to speak: since she has enough beads of each color to select all $20$ from a single color, the number of ways to choose the $20$ beads is $$\binom{20+n-1}{n-1}=\binom{19+n}{n-1}=\binom{19+n}{20}\;.$$ (The linked article has a reasonably good explanation of this formula.) You therefore have

$$\binom{19+n}{20}=230,230\;.$$

A little trial and error reveals that $230,230=\dbinom{26}{20}$, so $19+n=26$, and $n=7$.

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+Nice answer, thorough, and quick(er than me)! – amWhy Jan 28 at 2:37
@amWhy: Thanks. I was hoping to find a nice way to avoid the trial and error, but nothing occurred to me quickly. – Brian M. Scott Jan 28 at 2:38
Why do you have 20 as the bottom of the third equation in the top? I don't get how $n-1$ turned into $20$, even after reading Wikipedia. – Doug Smith Jan 28 at 3:41
@Doug: I’m just using the fact that $\binom{m}k=\binom{m}{m-k}$, here with $m=19+n$ and $k=n-1$, so that $m-k=20$. – Brian M. Scott Jan 28 at 3:44
Ah. Thank you for the fantastic explanation. – Doug Smith Jan 28 at 4:22
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