Final answer:
To solve for the number of distinct combinations of three bills from $1, $5, $10, and $20 bills, we use the combination formula. After accounting for all possible groups of bills, including those with repeats, there are a total of 20 distinct groups, making option (A) the correct answer.
Step-by-step explanation:
The question involves calculating the number of distinct groups of 3 bills that can be made using $1 bills, $5 bills, $10 bills, and $20 bills. This is a problem of combination where we need to select 3 out of the 4 types of bills, without regard to the order. To solve this, we use the combination formula which is C(n, k) = n! / (k!(n - k)!) where 'n' is the total number of items, 'k' is the number of items to choose, '!' denotes the factorial of a number.
There are 4 types of bills, so n = 4, and we want to pick groups of 3, so k = 3. Calculating C(4, 3) gives us 4! / (3!(4 - 3)!) = (4*3*2*1) / (3*2*1) = 4. This means there are 4 ways to choose 3 different types of bills without regard to order, but we also need to consider combinations with repeated bills, for example, three $1 bills or two $1 bills and one $5 bill.
The different groups are: {1,1,1}, {5,5,5}, {10,10,10}, {20,20,20}, {1,1,5}, {1,1,10}, {1,1,20}, {5,5,1}, {5,5,10}, {5,5,20}, {10,10,1}, {10,10,5}, {10,10,20}, {20,20,1}, {20,20,5}, {20,20,10}, {1,5,10}, {1,5,20}, {1,10,20}, {5,10,20}. This accounts for 20 distinct groups. Thus, the correct answer is (A) 20.