Question
How many four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?

Answer

Since the required num bars are greater than 5000.
$\therefore$ the thousand's place can be filled with any of two digits 5 or 9.
So, there are 2 ways of filling the thousand's place.
Since repetition of digits is not allowed, so the hundred's ten's and one's places can be filled in 4, 3 and 2 ways respectively.
Hence, the required number of num bers = 2 × 4 × 3 × 2 = 49

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