Question

An old-fashioned Chinese restaurant offers a family dinner where you get to choose one dish from “column A” (which has 8 dishes), one dish from “column B” (which has 10 dishes) and one dish from “column C” (which has 5 dishes). How many different family dinners can be chosen?

Answer 1

The answer is 10 times 8 times 5 or 400

Answer 2

Answer: 400

Explanation:

Given : The number of dishes in column A = 8

The number of dishes in column B = 10

The number of dishes in column C = 5

Since for dinner we need to choose one dish from each column, then

The total number of combinations of dishes for dinner will be :-

`8*10*5=400`

Hence, the number of family dinners =400