Question

How many differnet numbers, greater than 50000 can be formed with the digits 0, 1,1,5,9.

Answer 1

Total number of digits = 5

Now, numbers greater then 50000 will have either 5 or 9 in the first palce and will consist of 5 digits.

Number of numbers of which digit 5 at first palce =$\frac{4!}{2!}$ [$\because$ 1 is repeated]

= $\frac{4\times 3\times 2!}{2!}$

= 12

Number of numbers with digit 9 at first place

= $\frac{4!}{2!}=12$

Hence, the required number of numbers

= 12+12= 24.