Question

Quality control inspector selects two parts out of five for inspecting defects. How many permutations can be seleced?

Answer 1

Answer:

P(n,r) = 20

Step-by-step explanation:

We given that

The inspector have to select from five parts

So

n = 5

And

He selected parts = 2

So

r = 2

We also know that

permutation of n and r = P(n,r) = n!/(n-r)!

Putting values we get

P(n,r) = 5! / (5-2)! = 5! / 3! = ( 5×4×3! ) / 3! = 5×4 = 20

Thus

P(n,r) = 20

There for there are 20 permutations.