a) In the store "Everything for Tea," there are 5 different cups and 3 different saucers. How many ways can you buy a cup and a saucer?

b) The store also has 4 teaspoons. How many ways can you buy a set consisting of a cup, a saucer, and a teaspoon?

c) The store still sells 5 cups, 3 saucers, and 4 teaspoons. How many ways can you buy two items with different names?
a) Let's choose a cup. To it, we can choose any of the three saucers. Therefore, there are 3 different sets containing the chosen cup. Since there are 5 cups in total, the number of different sets is 5$\cdot$3 = 15.
b) Any of the 15 sets from part a) can be complemented with a spoon in 4 ways. Therefore, the total number of possible sets is $15 \cdot 4=60$.

c) There are three different cases: the first - a cup with a saucer, the second - a cup with a spoon, the third - a saucer and a spoon. In each of these cases, it is easy to count the number of possible options (15 in the first, 20 in the second, 12 in the third). Adding them up, we get the total number of possible options: 47.

## Answer

a) 15;
b) 60;
c) 47 ways.