**Problem: **Determine the number of 3 digit numbers in base 10 having at least one 5 and at most one 3.

**Discussion:**

*(Suggested by Shuborno Das in class)*

From 100 to 999 let us count the number of numbers without the digit 5. We have 8 choices for first digit (can’t use 0 or 5), and 9 choices each for second and third spot (skipping the digit 5 in each case). Hence \(900 – 8 \times 9 \times 9 = 252 \) three digit numbers have at least one 5.

From these 252 numbers, we must delete the numbers which have more than one 3. Since the number already has at least one 5, it may have at most two 3’s. The possible numbers that can be constructed by one 5 and two 3’s are \(\frac{3!}{2!} = 3 \).

Hence the numbers with at least one 5 and at most one 3 are 252 – 3 = 249.

## Chatuspathi:

**Paper:**RMO 2015 Mumbai Region**What is this topic:**Combinatorics**What are some of the associated concepts:**Complimentary Counting**Where can learn these topics:**Cheenta**Book Suggestions:**Principles and Techniques in Combinatorics