# Simple as that

**Discrete Mathematics**Level 4

Consider three **natural numbers** \(n_{1} ,n_{2}, n_{3}\) defined as

\(- n_{1`}=x_{1}x_{2}x_{3}x_{4}\) ( \(4~digit~number\))

\(- n_{2}=y_{1}y_{2}y_{3}\) (\(3~digit~number\))

\(- n_{3}=z_{1}z_{2}\) (\(2~digit~number\))

where \(x_{1}....x_{4},y_{1}...y_{3},z_{1},z_{2}\) are **all non zero digits**.

Then total number of **nine digit numbers** formed by picking up the digits used in the numbers \(n_{1},n_{2},n_{3}\) if digits are taken only from the **right side** of either of the numbers \(n_{1},n_{2}~and~n_{3}\)