Question

Find the number of 3-digit numbers in which the digit at the hundredth place is greater than the other two.

Answer 1

Hint: To solve this question, we need to understand that is the hundredth place, tens place and unit place, i.e. the concept of place value. Then we will find all the possible combinations by starting with the largest digit in the hundred’s place. Then, we will go down to second largest and so on. We will add all the combinations to find the number of 3-digit numbers in which digit at the hundredth place is greater than the other two.

Complete step-by-step answer:
In a number, the digit at the right most position is known as a digit on unit’s place. Left of unit’s place is ten’s place. The value of a digit is 10 times the same digit at unit’s place. The digit at the left of ten’s place is hundredths place and the value of digit at hundreds place is 100 times the value of the same digit at unit’s place and it is 10 times the value of the same digit at the tens place. This concept is known as place value of digits.
It is given that the number we have to form is a 3-digit number. Therefore, the right most position will be the unit's place, then left of it is ten’s place and left of it and at left most, we have hundreds place.
Now, the digit at hundred’s place must be greater than the other two.
Suppose we have 9 at hundreds place. Therefore, the number will be 9 _ _. Any digit smaller than 9 can occupy those space. There are 9 such digits.
So, ten’s place has 9 chances and the unit's place has 9 chances. Thus, the number of numbers will be 81.
Now, if the number at hundreds place is 8. Other two spaces will have digits smaller than 8. Thus, each space has 8 chances. Therefore, the number of such numbers is 64.
Similarly,
7 _ _ = 49
6 _ _ = 36
5 _ _ = 25
4 _ _ = 16
3 _ _ = 9
2 _ _ = 4
1 _ _ = 1
Therefore, the number of numbers = 81 + 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 285.

Note: It is to be noted that the number of numbers such that the digit at hundreds place is greater than the other two are equal to the square of the digit at the hundred’s place. Students can directly remember that the sum of squares of first n integers is given as $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.