In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included? Choose the correct choice:  A) 36  B) 51  C) 70  D) 87  E) 101
I. Case I = Four Quarters = this is impossible, because if we already have a dollar in quarters, then we either have to have one dime or nickels &amp; pennies. We can't have all four coins represented if we have four quarters.   II. Case II = Three Quarters (75 cents)   This allows for two or one dime. Three dimes would bring us up to $1.05, and we wouldn't have room for both pennies &amp; nickels.   Subcase 1 = 3Q, 2 D (95 cents)   Here, we could have have   1) two nickles   2) one nickle   (Notice that once the number of Q &amp; D &amp; N are determined, the number of pennies is fixed and doesn't have to be considered.   Subcase 2 = 3Q, 1 D (85 cents)   Here we could have   3) 4 nickles   4) 3 nickles   5) 2 nickles   6) 1 nickle   Case II allows for a total of six cases.   III. Case III = Two Quarters (50 cents)   This allows for 1-5 dimes.   Subcase 1 = 2Q, 5 D = we could have 1 nickle (1 case)   Subcase 2 = 2Q, 4 D = we could have 1-3 nickles (3 cases)   Subcase 3 = 2Q, 3 D = we could have 1-5 nickles (5 cases)   Subcase 4 = 2Q, 2 D = we could have 1-7 nickles (7 cases)   Subcase 5 = 2Q, 1 D = we could have 1-9 nickles (9 cases)   Case III allows for a total of 25 cases   IV. Case IV = One Quarter (25 cents)   This allows for 1-7 dimes   Subcase 1 = 1 Q, 7D = we could have 1-2 nickles (2 cases)   Subcase 2 = 1 Q, 6D = we could have 1-4 nickles (4 cases)   Subcase 3 = 1 Q, 5D = we could have 1-6 nickles (6 cases)   Subcase 4 = 1 Q, 4D = we could have 1-8 nickles (8 cases)   Subcase 5 = 1 Q, 3D = we could have 1-10 nickles (10 cases)   Subcase 6 = 1 Q, 2D = we could have 1-12 nickles (12 cases)   Subcase 7 = 1 Q, 1D = we could have 1-14 nickles (14 cases)   Case IV allows for a total of 56 cases.   There's no other case, because we have to have at least one quarter. The total over the cases equals   Total = 6 + 25 + 56= 
<gadget id="calculator">6 + 25 + 56</gadget>
<output>87</output>
87   ANSWER:87
<result>D</result>
In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?