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

Question:

A child has three plastic toys bearing the digits 3, 3, 5 respectively. How many 3 - digit numbers can he make using them?

 

Solution:

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

 

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