Someone shot 10 arrows at a target with 10 concentric rings, each being worth a different integer number of points from 1 to 10. How many different ways are there of scoring 10 points by doing this? (Note that not all the arrows have to hit the target and that order matters; 6 first then 4 is different from 4 first then 6. Also note that two or more arrows may hit the same ring.)

(In reply to re: Intuitive Representation IMHO by David Shin)

Mmmmm, I think it’s ok the way I set up the dividers. I’m not sure if you read through my whole comment, but two of the dividers are fixed, so I can only choose where 9 of the dividers go. So I think we are saying the same thing, I am just using two dividers as place holders. If I didn’t have the two dummies, I could have some problems, like only making 8 or 9 bins or something.

I didn’t think it was hard to count how many of the above configurations there were. If you name the slots a through k, with a being before the first ball, and k being after the last ball, what I am basically doing is saying for example, "Ok, I will put my dividers in slots c, k, e, h, b, e, k, i, e." Since I have 11 slots that I can pick from to put my 9 dividers, that is 11 choose 9. Since the order in which I list the slots where I am putting the dividers doesn’t matter (the outcome of the arrangement will be the same either way I list it), that means I should count combinations, not permutations. And since I can pick a slot more than once (like e or k in my example), that means that replacement is allowed.

So for combinations with replacement, the formula for 11 choose 9 is (11+9-1)!/[9!*(11-1)!]. This turns out to be 19!/(9!*10!) = 92378, which is the same as what you got when you went another way and counted 19 choose 10 combinations WITHOUT replacement.

Posted by nikki on 2004-09-24 08:40:30