<h2>Explanation</h2><h2></h2><h2>He can use the 2 pennies in 3 ways, take none, take 1 or take 2</h2><h2>He can use the 3 nickels in 4 ways, take none, take 1, etc</h2><h2>the dime in 2 ways,</h2><h2>and the quarters in 3 ways</h2><h2 /><h2>total number of ways = (3)(4)(2)(3) = 72</h2><h2>but that includes taking none of any coins,</h2><h2 /><h2>so number of ways to use the coins is 71</h2><h2 /><h2>Here is a little silly computer program that finds and lists all the possible sumes</h2><h2 /><h2>10 for p = 0 to 2</h2><h2>20 for n = 0 to 3</h2><h2>30 for d = 0 to 1</h2><h2>40 for q = 0 to 2</h2><h2>50 sum = p+5*n+10*d+25*q</h2><h2>60 print sum,</h2><h2>70 next q</h2><h2>80 next d</h2><h2>90 next n</h2><h2>100 next p</h2><h2>>run</h2><h2>0 25 50 10 35 60 5 30 55 15 40 65 10 35 60 20</h2><h2>45 70 15 40 65 25 50 75 1 26 51 11 36 61 6 31</h2><h2>56 16 41 66 11 36 61 21 46 71 16 41 66 26 51 76</h2><h2>2 27 52 12 37 62 7 32 57 17 42 67 12 37 62 22</h2><h2>47 72 17 42 67 27 52 77</h2><h2 /><h2>Notice there are duplications,</h2><h2>e.g. picking 2 nickels or 1 dime yields the same sum</h2><h2 /><h2>All you have to do is eliminate all duplications, and not count the 0</h2><h2>I think there are 47 or them</h2>
Irrational. though 25 can be square rooted to find an integer, 35 cannot. the solution would be 0.84515425... and is irrational.
60% chance for red and 40% for silver. you are most likely to get a red than a silver
