Question

A ten digit number is formed using the digit from 0 to 9, every digit being used exactly once. Find the probability that the number is divisible by 4.

• A. $20/81$
• B. $16/81$
• C. $22/81$
• D. $25/81$

Answer 1

The correct option is A $20/81$

The digits are $0,1,2,\mathrm{3..9}$
Number divisible by $4$ have the last two digits such that the number formed by them is divisible by $4$
These are either $04,08,20,40,60,80$

i.e in all and $6(8!)$ number can be formed with them.
Those which do not have $0$ in the last two digits and are divisible by $4$ and $16$ in numbers as $44$ and $88$ cannot be considered
Hence they are $16(8!-7!)$ in all.
Hence favorable ways are $6(8!)+16(8!-7!)$
Total numbers $=10!-9!$
Required probability $=\frac{6.8!+16(8!-7!)}{10!-9!}=\frac{20}{81}$