Depth · one concept from first principles · step 2 of 4
Mia has the coins shown below. To buy one toy that costs ¢, find how many different ways she can pay exactly the price of the toy.
| Dimes (10¢) | Nickels (5¢) | Pennies (1¢) |
|---|---|---|
| 2 | 2 | 10 |
Turn each coin into cents — dime 10, nickel 5, penny 1 — and hunt combinations totaling 20 cents within her supply.
Turning every coin into its cent value makes the total easy to add up and compare to 20.
2.MD.C.8Analyze The UnitsTwo dimes already make 20 cents, so no nickels or pennies are needed — that's 1 way.
Two dimes hit 20 exactly, leaving nothing more to add.
2.MD.C.8Make A Systematic ListOne dime leaves 10 cents: 2 nickels; 1 nickel + 5 pennies; or 10 pennies — all within her limits, so 3 ways.
After one dime, the leftover 10 cents can be split among nickels and pennies in a few neat ways.
2.MD.C.8Make A Systematic ListWith no dimes, only 2 nickels + 10 pennies reach 20 cents; fewer nickels would need 15 or 20 pennies she lacks — 1 way.
With only 10 pennies on hand, she must use both nickels, leaving exactly one workable combination.
2.MD.C.8Make A Systematic ListAdd the cases — 1 with 2 dimes, 3 with 1 dime, 1 with 0 dimes: 1 + 3 + 1 = 5.
The cases by number of dimes don't overlap, so the total is just their sum.
2.MD.C.8Make A Systematic ListThe total number of ways is the sum of the ways counted in each separate case grouped by how many dimes are used.
Sorting every payment by how many dimes it uses breaks all the ways into three groups that miss nothing and share nothing, so the group counts simply add up to the total.
A payment can use only 0, 1, or 2 dimes since Mia has just 2 dimes, and each payment has one fixed number of dimes, so every way falls into exactly one of the three groups with none left out and none counted twice.
Organize your guesses by how many dimes you use, and the rest of the coins fall into place by twenties!
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