Question

A ten digit number is formed using the digit from $0$ to $9$, every digit being used exactly once. The probability that the number is divisible by $4$ is $5k/81$.Find the value of $k$.

Answer 1

Let $n$ = total number of ways $=10!-9!$
To find the favorable number of ways. we observe that a number is divisible by 4 is the last two-digit number is divisible by 4. Hence, the last two digitd can be $20,40,60,80,12,32,52,72,92,04,24,64,84,16,36,56,76,96,08,28,48,68.$
Corresponding to each of $20,40,60,80,04,08,$ the remaining 8 places can be filled in $8!$ ways so that the number of ways in this case $=6\times 8!$
And corresponding to remaining $16$ possibilities, the number of ways $16(8!-7!)$
Hence, $m=$ favorable number of ways $=6\times 8!+16(8!-7!)=22\times 8!-16\times 7!$
Therefore the required probability $=\frac{m}{n}=\frac{22\times 8!-16\times 7!}{10!-9!}=\frac{160}{648}=\frac{20}{81}$
$\therefore k=4$