How many different dimes can be made with 1, 2, and 5 cents? Coins can be reused.

This is a direct solution. The basic idea is to classify all the solutions according to the individual rescue of 1 nickel.

Suppose there are 20 nickels and there is no dichotomy, so there is only one way. Suppose there are 19 nickels, and the monetary value is 5× 19 = 95 cents, then the total monetary value should not exceed 1 yuan = 100 cents, and the monetary value of the taken two cents should not exceed 5 cents. Obviously,

1+3+6+8+ 1 1+ 13+……+48+5 1

=( 1+48)+(3+46)+(6+43)+……+(23+26)+5 1

=49× 10+5 1

=54 1 (species)

Answer: * * * There are 54 1 combinations.

Solution 2 This is a smart and simple algorithm.

Divide 50 binary coins and 20 nickels into two groups A and B. Because the total currency value of these coins is 50×2+20×5=200 (cents), the currency values of the two groups are nothing more than the following three situations.

(1) The money in group A is less than 1 yuan, and the money in group B is more than1yuan.

(2) Group A has one yuan more money and Group B has less money 1 yuan.

(3) The money in Group A and Group B is equal, which is one yuan.

There are two points to pay special attention to here: First, case (1) and case (2) are symmetrical, but A and B are interchanged. Second, all the possibilities of (1) plus all the possibilities of (3) are the answers to our questions.

What are (1) and (3)?

First, calculate how many different ways to group the above * * * totals. Because there are 50 binary coins, there are 5 1 ways to divide them. Similarly, there are 20 nickels, so there are 2 1 ways to divide them. Therefore, there are always * * *, and there are 2 1 division methods.

Let's take a look at how many ways to divide the money in group A and group B when they are all one yuan. Obviously, there must be an even number of nickels at this time (why? ), so the number of nickels can be 0, 2, ..., 20 and * * * can be divided into eleven ways.

According to the symmetry of case (1) and case (2), it is easy to know the number of (1× 511) ÷ 2 = 530.

The number of (1) plus the number of (3) is 530+11= 541(species). This is the answer.

This is a problem that combines thinking and calculation. Many students use the scheme of 1. But most of them didn't understand correctly. It may be unclear. Learning "number" is the basic skill of mathematics, and we can't be careless. To improve your "number" ability, try another method.