Carla has 10 coins, each of which is a nickel, a dime, or a quarter, or a penny. The total value of her coins is less than $1.00. How many different combinations of coins might Carla have?
+592 Since the total is less than a dollar, dimes < 7, quarters < 4
--> 4P + 2N + 1D + 3Q
--> 5P + 1N + 1D + 3Q
--> 3P + 1N + 4D + 2Q There are 6 combinations for when there is 1 < Q < 4
--> 4P + 1N + 3D + 2Q
--> 3P + 2N + 3D + 2Q
--> 2P + 3N + 3D + 2Q
--> 2P + 1N + 6D + 1Q
--> 1P + 2N + 6D + 1Q
--> 3P + 1N + 5D + 1Q
--> 2P + 2N + 5D + 1Q
--> 1P + 3N + 5D + 1Q
--> 4P + 1N + 4D + 1Q
--> 3P + 2N + 4D + 1Q. There are 2 + 3 + 4 + 5 + 6 + 7 = 27 combinations when there is 1 quarter
--> 2P + 3N + 4D + 1Q
--> 1P + 4N + 4D + 1Q
...
--> 7P + 1N + 1D + 1Q
--> 6P + 2N + 1D + 1Q
--> 5P + 3N + 1D + 1Q
--> 4P + 4N + 1D + 1Q
--> 3P + 5N + 1D + 1Q
--> 2P + 6N + 1D + 1Q
--> 1P + 7N + 1D + 1Q
This means the total combination is 6 + 27 = 33 combinations, which is your answer :)
a=1;p=0; b=1;c=1;d=1;n=a*1+b*5+c*10+d*25;if(n<100 and (a+b+c+d)==10, goto loop, goto next);loop:p=p+1;printp," =",a,b,c,d;next:a++;if(a<60, goto5,0);a=1;b++;if(b<60, goto5, 0);a=1;b=1;c++;if(c<60, goto5,0);a=1;b=1;c=1;d++;if(d<60, goto5, discard=0;printp
p n d q
1 = 7 1 1 1
2 = 6 2 1 1
3 = 5 3 1 1
4 = 4 4 1 1
5 = 3 5 1 1
6 = 2 6 1 1
7 = 1 7 1 1
8 = 6 1 2 1
9 = 5 2 2 1
10 = 4 3 2 1
11 = 3 4 2 1
12 = 2 5 2 1
13 = 1 6 2 1
14 = 5 1 3 1
15 = 4 2 3 1
16 = 3 3 3 1
17 = 2 4 3 1
18 = 1 5 3 1
19 = 4 1 4 1
20 = 3 2 4 1
21 = 2 3 4 1
22 = 1 4 4 1
23 = 3 1 5 1
24 = 2 2 5 1
25 = 1 3 5 1
26 = 2 1 6 1
27 = 1 2 6 1
28 = 6 1 1 2
29 = 5 2 1 2
30 = 4 3 1 2
31 = 3 4 1 2
32 = 2 5 1 2
33 = 1 6 1 2
34 = 5 1 2 2
35 = 4 2 2 2
36 = 3 3 2 2
37 = 2 4 2 2
38 = 1 5 2 2
39 = 4 1 3 2
40 = 3 2 3 2
41 = 2 3 3 2
42 = 3 1 4 2
43 = 5 1 1 3
44 = 4 2 1 3
