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

1. A

$\frac{5040}{9999}$

2. B

$\frac{5000}{9999}$

3. C

$\frac{5030}{9999}$

4. D

None of these

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### Solution:

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

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