Question

In a lottery 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

Please note: nPr = n! / (n - r)!

nCr = n! / [(n - r)! * r!]

0 is not included as 0 is neither positive nor negative.

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(a) Total Number of different outcomes

Arrange 5 numbers out of 90= 90P5 = 90! / (90-5)! = 90! / 85! = 90 * 89 * 87 * 86 * 85 = 5,273,912,160

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(b) Total Outcomes

Choose 4 numbers out of 89 (Since 1 is already there) and arrange them in 5! ways

= 90C4 * 5! = 306,622,800

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(c) There are 49 numbers below 50 and 30 numbers above 60

Therefore Possible outcomes, choose 2 numbers from 1 - 49 and choose 3 numbers from 61 - 90 and then arrange them in 5! ways.

49C2 * 30C3 * 5! = 572,947,200

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(d) We have 10 single digit numbers from 0 till 9.

From 1 - 90: 10 numbers end with 1, 10 numbers end with 2, 10 end with 3 and so on....

Therefore out of 10 different last digits, we choose 5 and then arrange them in 5! ways = 10C5 * 5! = 252 * 120 = 30,240

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