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 $1.05 (105 cents), if at least one coin of at least three types of coins must be included?  Choose the correct choice from the following options.
A) 78  B) 103  C) 118  D) 131  E) 221

Solution: This question is similar to determining the number of ways using one of each type of coin, but allows the exclusion of one type.   I. Case I = Four Quarters (100 cents)   Subcase 1= 4 Q, 1 D (110 cents) not allowed as this exceeds $1.05.   Subcase 2= 4 Q, 0 D (100 cents) not allowed since it does not allow for both nickels and pennies.   Case I does not allow any ways.   II. Case II = Three Quarters (75 cents)   This allows for two, one, or zero dimes. Three dimes would bring us up to $1.05, and we wouldn't have room for any pennies or nickels.   Subcase 1 = 3 Q, 2 D (95 cents)   1) one nickel (N) and 5 pennies (P)   2) 2 N and 0 P   3) 0 N and 10 P   Subcase 2 = 3 Q, 1 D (85 cents)   4) 4 N and 0 P   5) 3 N and 5 P   6) 2 N and 10 P   7) 1 N and 15 P   8) 0 N and 20 P   Subcase 3 = 3 Q, 0 D (75 cents)   (5 ways) we could have 1-5 nickels   Case II allows for a total of 13 ways.   III. Case III = Two Quarters (50 cents)   This allows for 0-5 dimes.   Subcase 1 = 2 Q, 5 D (100 cents)   (2 ways) we could have 0-1 nickel   Subcase 2 = 2 Q, 4 D (90 cents)   (4 ways) we could have 0-3 nickels   Subcase 3 = 2 Q, 3 D (80 cents)   (6 ways) we could have 0-5 nickels   Subcase 4 = 2 Q, 2 D (70 cents)   (8 ways) we could have 0-7 nickels   Subcase 5 = 2 Q, 1 D (60 cents)   (10 ways) we could have 0-9 nickels   Subcase 6 = 2 Q, 0 D (50 cents)   (10 ways) we could have 1-10 nickels   Case III allows for a total of 40 ways   IV. Case IV = One Quarter (25 cents)   This allows for 0-7 dimes   Subcase 1 = 1 Q, 7 D (95 cents)   (3 ways) we could have 0-2 nickels   Subcase 2 = 1 Q, 6 D (85 cents)   (5 ways) we could have 0-4 nickels   Subcase 3 = 1 Q, 5 D (75 cents)   (7 ways) we could have 0-6 nickels   Subcase 4 = 1 Q, 4 D (65 cents)   (9 ways) we could have 0-8 nickels   Subcase 5 = 1 Q, 3 D (55 cents)   (11 ways) we could have 0-10 nickels   Subcase 6 = 1 Q, 2 D (45 cents)   (13 ways) we could have 0-12 nickels   Subcase 7 = 1 Q, 1 D (35 cents)   (15 ways) we could have 0-14 nickels   Subcase 8 = 1 Q, 0 D (25 cents)   (15 ways) we could have 1-15 nickels   Case IV allows for a total of 78 ways.   IV. Case V = Zero Quarters (0 cents)   This allows for 1-9 dimes   Subcase 1 = 0 Q, 9 D (90 cents)   (2 ways) we could have 1-2 nickels   Subcase 2 = 0 Q, 8 D (80 cents)   (4 ways) we could have 1-4 nickels   Subcase 3 = 0 Q, 7 D (70 cents)   (6 ways) we could have 1-6 nickels   Subcase 4 = 0 Q, 6 D (60 cents)   (8 ways) we could have 1-8 nickels   Subcase 5 = 0 Q, 5 D (50 cents)   (10 ways) we could have 1-10 nickels   Subcase 6 = 0 Q, 4 D (40 cents)   (12 ways) we could have 1-12 nickels   Subcase 7 = 0 Q, 3 D (30 cents)   (14 ways) we could have 1-14 nickels   Subcase 8 = 0 Q, 2 D (20 cents)   (16 ways) we could have 1-16 nickels   Subcase 9 = 0 Q, 1 D (10 cents)   (18 ways) we could have 1-18 nickels   Case V allows for a total of 90 ways.There's no other case, because we have to have at least one quarter or one dime. The total over the cases equals   Total = 0 + 13 + 40 + 78 + 90= 
0 + 13 + 40 + 78 + 90 = 221
221 ways.   OA=(221)
Answer: E