Option 2 : ^{9}P_{4}

**Given**:

4 unique balls are put into 9 unique boxes if no box can contain more than one ball

**Concept used:**

Permutations

**Formula used:**

^{n}P_{r }= n!/(n - r)!

**Calculation:**

1^{st} ball can be placed into any of the 9 boxes.

Since no box can contain more than one ball, we have 8 boxes remaining.

2^{nd} ball can be placed into any of these 8 boxes.

Similarly, 3^{rd} ball can be placed into any of the remaining 7 boxes.

4^{th} ball can be placed into any of the remaining 6 boxes.

Hence, the required number of ways = ^{9}P_{4}

**∴ The required number of ways is ^{9}P_{4}**.