There are 9999 tickets bearing numbers 0001, 0002, … , 9999. if one ticket is selected from these tickets at random, the probability that thenumber on the ticket will consists of all different digits, is

There are 9999 tickets bearing numbers 0001, 0002, ... , 9999. if one ticket is selected from these tickets at random, the probability that thenumber on the ticket will consists of all different digits, is

A
$\frac{5040}{9999}$
B
$\frac{5000}{9999}$
C
$\frac{5030}{9999}$
D
None of these

Solution:

Total number of cases = 9999
Favorable cases = 10 x 9 x8 x 7 = 5040
Probability $= \frac{5040}{9999}$

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