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?

 400

Step-by-step 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 :-

Hence, the number of family dinners =400

step-by-step explanation:

i think the answer is b