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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?

asked
User Brown A
by
8.3k points

1 Answer

2 votes

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* ^1C_1


\text{Total ways}=(9!)/(3!(9-3)!)* 1


\text{Total ways}=(9* 8* 7* 6!)/(3* 2* 1* 6!)


\text{Total ways}=(9* 8* 7)/(6)


\text{Total ways}=3* 4* 7


\text{Total ways}=84

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

answered
User Timr
by
8.6k points

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