How many positive three-digit integers have an odd digit in both the

Hi SW4,

This is a typical permutation problem where the method of counting comes in very handy.

The following are the steps you need to follow while using the counting method.
1. Make empty spaces and try to fill up the empty spaces one by one
2. If a condition is given, you always find the condition space and fill the condition space first.

Here we need to form a 3 digit number given 2 conditions that the tens and units digit are odd. the digits we can use here are all the digits from 0 to 9 and since nothing is mentioned about the digits being repeated or not, we always consider the with repetition case.

So making 3 empty spaces _ Odd Odd.

The total number of ways of filling the units digit is 5, since we can use any one out of 1,3,5,7 and 9. Similarly the total number of ways of filling the tens digit is 5. Now since the digits can be repeated and we need to have a 3 digit number, we can fill the hundreds digit in 9 ways, since we can use any one out of 1,2,3,4,5,6,7,8,9. We cannot use 0 since we need a 3 digit number.

So the answer here will be 9 * 5 * 5 = 225

Hope this helps!

CrackVerbal Academics Team