How many different ways can you make 50 cents

Question

How many different ways can you make 50 cents if you have 1 quarter, 5 dimes, and 10 nickels?

Answer 1

The number is small enough here that you can just list the ways and count.

Note that the restriction to 1 quarter eliminates only one of the possibilities that we would otherwise have (5 dimes = 10 nickels equals 50 cents).

Writing them as (Quarters, Dimes, Nickels), we have

(0, 0, 10), (0, 1, 8), (0, 2, 6), (0, 3, 4), (0, 4, 2), (0, 5, 0), (1, 0, 5), (1, 1, 3), (1, 2, 1)

There are 9 ways.

Answer 2

Example: ₅C₂ translates to 5!/2!/(5-2)! or (5×4×3×2×1) ÷ (2×1) ÷ (3×2×1) or 120 ÷ 2 ÷ 6 equal to 10.

First use all 10 nickels. This is Solution A.

Next use all 5 dimes. This is Solution B.

Also include:

4 dimes and 2 nickels (Solution C) or ₅C₄ × ₁₀C₂ or 5 × 45 or 225.

3 dimes and 4 nickels (Solution D) or ₅C₃ × ₁₀C₄ or 10 × 210 or 2100.

2 dimes and 6 nickels (Solution E) or ₅C₂ × ₁₀C₆ or 10 × 210 or 2100.

1 dime and 8 nickels (Solution F) or ₅C₁ × ₁₀C₈ or 5 × 45 or 225.

Finally use the 1 quarter along with: (a) 2 dimes and 1 nickel; (b) 1 dime and 3 nickels; or (c) 5 nickels.

Now write ∑_{(Solution A through Solution F)} + [1 quarter × ₅C₂ × ₁₀C₁] + [1 quarter × ₅C₁ × ₁₀C₃] + [1 quarter × ₁₀C₅].

This is simplified to 1 + 1 + 2(225) + 2(2100) + [1 × 5!/2!/(5-2)! × 10!/1!/(10-1)!] + [1 × 5!/1!/(5-1)! × 10!/3!/(10-3)!] + [1 × 10!/5!/(10-5)!] which condenses to 4652 + [1 × 10 × 10] + [1 × 5 × 120] + [1 × 252] or 4652 + 100 + 600 + 252 or 5604 different ways.

If one only counts the possible coin combinations from 1 quarter, 5 dimes, and 10 nickels that will sum to 50 cents without distinguishing which of the dimes and nickels are being used, then the number of possible coin combinations does come to 9.