Question

Find the number of three digit numbers from 100 to 999 including all the numbers which have any one digit that is the average of the other two.

Answer 1

In this question the concept of permutations and combinations will be used. We will use logic to define all the cases possible, and then count the number of three digit numbers which have one digit as the average of the others. In order to select r unique things from a total number of n things, the number of ways are-
${}_{}^{\text{n}}{\text{C}}_{\text{r}}^{} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!{\text{r}}!}}$

Complete step-by-step answer:
We know that for the average of two numbers to exist, their sum should be an even number, this is because if the sum is odd, the average will be a fraction, which is not possible. So, we will form different cases for different averages and count the numbers in each case. Let-
O = {1, 3, 5, 7, 9} be a set of one digit odd natural numbers.
E = {0, 2, 4, 6, 8} be a set of one digit even natural numbers.

For the average to exist, either both the numbers are odd, or both are even. First from the set O, we will find the number of ways to select any two digits, which can be calculated as-
${}_{}^5{\text{C}}_2^{} = \dfrac{{5!}}{{3!2!}} = \dfrac{{120}}{{6 \times 2}} = 10$
Now these three digit numbers can be rearranged in 3! ways, hence the total numbers formed are-
10(3!) = 60 ways

Now, from the set E, we need to select two out of the four digits. We will not include 0 because it will add some extra cases. So we can select two digits as-
${}_{}^4{\text{C}}_2^{} = \dfrac{{4!}}{{2!2!}} = \dfrac{{24}}{{2 \times 2}} = 6$
Now these three digit numbers also can be rearranged in 3! ways, hence the total numbers formed are-
6(3!) = 36 ways

Now, we will consider the cases which have the digit 0. So, there are 4 ways to select one of the remaining four digits. Also, we know that 0 cannot be placed in the hundreds place, because we need to find only three digit numbers. So, the number of ways to fill the hundreds place are 2. The number of ways to fill the tens place and ones place are 2 and 1 respectively.
So the total number of ways are = 4 x 2 x 2 = 16 ways

Till now, we only selected unique digits. Now we will also consider the cases where all the digits are the same. Clearly 9 such numbers are possible(111, 222 and so on). On adding all these cases we get-
60 + 36 + 16 + 9 = 121 numbers.

Hence, three digit numbers from 100 to 999 including all the numbers which have any one digit that is the average of the other two are 121.

Note: In such types of questions, we need to analyze the problem and find all the cases that are possible. Alternatively, we can take each and every average possible(1, 2, 3 and so on) and then count the numbers possible for each average. This is a very long and tedious method. Therefore, we should always use the method of combinations and permutations.