Question

In a lottery of 5 different numbers are chosen from the first 90 positive integers.

a) How many possible outcomes are there? (An outcome is an unordered sample of five numbers)

b)How many outcomes are there with the number 1 appearing among the five chosen numbers?

c) How many outcomes are there with two numbers below 50 and three numbers above 60?

d)How many outcomes are there with the property that the last digits of all five numbers are different? (The last digit of 5 is 5 and the last digit of 34 is 4.)

Answer 1

A) Number of outcomes possible = 90C5

= 90! /(5! x 85!)

= 43,949,268

B) Number of oucomes with one number being number 1 = 1 x 89C4

= 1 x 89!/(4!x84!)

= 2,441,626

C) Number of oucomes with 2 numbers below 50 and three numbers above 60

= 49C2 x (90-60)C3

= 49C2 x 30C3

= 1176 x 4060

= 4,774,560

d) Number of outcomes with same last digit = 9

So, number of options available reduces by 9 on each selection of a number

So, total number of options available = 90 x 81 x 72 x 63 x 54 /5!

= 14,880,348