Question

We select two distinct numbers (a, b) in the range 1 to 99 (inclusive). How many ways can we pick a and b such that their sum is even and a is a multiple of 9?

Answer 1

Answer:

The possible number of ways to select distinct (a, b) such that (a + b) is even is 534.

Step-by-step explanation:

The range 1 - 99 has 99 numbers, since 1 and 99 are inclusive.

Of these 50 numbers are odd and 49 are even.

The two distinct numbers a and b must have an even sum and a should be a multiple of 9.

The sum of two numbers is even only when both are odd or both are even.

The possible values that a can assume are,

a = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99}

Thus, a can assume 6 odd values and 5 even values.

• If a = odd number, then b can be any of the 49 out of 50 odd numbers.

Total number of ways to select a and b such that both are odd and their sum is even is:

$n(Odd\ a\ and\ b)=n(Odd\ value\ of\ a)\times n(Odd\ value\ of\ b)=6\times49=294$

• If a = even number, then b can be any of the 48 out of 50 even numbers.

Total number of ways to select a and b such that both are even and their sum is even is

$n(Even\ a\ and\ b)=n(E\ value\ of\ a)\times n(Even\ value\ of\ b)=5\times48=240$

Total number of ways to select distinct (a, b) such that (a + b) is even is =

$=n(Odd\ a\ and\ b)+n(Even\ a\ and\ b)=294+240=534$

Thus, the possible number of ways to select distinct (a, b) such that (a + b) is even is 534.