Question

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

Answer 1

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.

$\text{Total ways}=^9C_3\times ^1C_1$

$\text{Total ways}=\dfrac{9!}{3!(9-3)!}\times 1$

$\text{Total ways}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}$

$\text{Total ways}=\dfrac{9\times 8\times 7}{6}$

$\text{Total ways}=3\times 4\times 7$

$\text{Total ways}=84$

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