A child has three plastic toys bearing the digits 3, 3, 5 respectively.

To find: number of 3 digit numbers he can make

If all were distinct, he could have made $3 !=6$ numbers

But 2 number are the same

So the number of possibilities $=\frac{3 !}{2 !}=\frac{6}{2}=3$

He can make 3 three - digit numbers using them