Browse Questions

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

$\begin {array} {1 1} (1)\;\large\frac{5040}{9999} & \quad (2)\;\large\frac{5000}{9999} \\ (3)\;\large\frac{5030}{9999} & \quad (4)\;None\: of \: these \end {array}$

The total number of tickets is 9999. Also four different digits can be
arranged in $10 \times 9 \times 8 \times 7 = 5040$ ways.
Hence the required probability is $\large\frac{5040}{9999}$
Ans : (1)
edited Sep 14, 2014