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I just finished it now, to see if I had been on the right track after all, and I got 242. Here's the approach I was using (no real algebra or recursion involved, sadly):<br /><br />I looked at it in terms of how many Quarters, or Q's you used. So..<br /><br />4 Q's has 1 combination only.<br /><br />3 Q's has combinations involving 2 Dimes (2 D's), 1 D, or 0 D. When you have 2 D's, you're making up the remaining 5 cents in combos of Nickels (N's) and Pennies (P's). Since choosing the value for N dictates the value for P, you really only have to figure out how many N combinations there are. And that number is essentially 0 through x, where 5x is less than or equal to the remaining value after Q's and D's. For 5 cents, there are only 2 combinations (0 N's, 1 N). <br /><br />When you have 3 Q's and 1 D, you're making up 15 cents in N's and P's, so there are 4 combinations (0 N's, 1 N, 2 N's, 3 N's). <br /><br />And when you have 3 Q's and 0 D's, you're making up 25 cents, so there are 6 combinations. <br /><br />Shortening my notation then:<br /><br />4 Q's = 1 combination<br /><br />3 Q's = (2 D's + 1 D + 0 D's) = 2 + 4 + 6 = 12<br /><br />2 Q's = (5 D's + 4 D's + ... + 0 D's) = 1 (no N's or P's) + 3 (for 10 cents) + 5 (for 20 cents) + 7 (for 30 cents) + 9 (for 40 cents) + 11 (for 50 cents) = 36<br /><br />1 Q's = (7 D's + 6 D's + ... + 0 D's) = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = 72<br /><br />0 Q's = (10 D's + 9 D's + ... + 0 D's) = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121<br /><br />So all combinations = 1 + 12 + 36 + 72 + 121 = 242<br /><br />When I was doing this the first time, I didn't hit on the fact that, when fixing the # of Q's, and then breaking down the possible # of D's for each, that I then only had to look at each increment of N's (since the # of P's is fixed by the # of N's). Instead, I stupidly thought I had cascading permutations, meaning that I had Q's fixed, then had to enumerate D's, and then N's within D's, and then P's within N's. Had I spent 10 more minutes following this path I might've even gotten the right answer.<br /><br />Oh well... I guess I'm not the math whiz I was 25 years ago!Kimota94 aka Matt aka AgileManhttps://www.blogger.com/profile/00404161474780005815noreply@blogger.com