In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one of the four you selected is defective?

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one  of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

Therefore, the four calculators can be selected in 84 ways.

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =