Question

If two different numbers are taken from the set $\{0,1,2,3,......,10\}$; then the probability that their sum as well as absolute difference are both multiple of $4$, is.

• A. $\frac{6}{55}$
• B. $\frac{12}{55}$
• C. $\frac{14}{45}$
• D. $\frac{7}{55}$

Answer 1

The correct option is A $\frac{6}{55}$

Let us take this case by case. First, take number $0$. The second number $p$ has to be chosen such that $p-0$ and $p+0$ are divisible by $4$. Hence, $p$ can be either $4$ or $8$.

Now, take number $1$. We find that $3+1$ is divisible by $4$ but $3-1$ is not. Similarly, $5-1$ is divisible by $4$ but $5+1$ is not. You can see that for all odd numbers, no such $p$ exists, so only even numbers should be checked.

First Number	Second Number	No. of cases
$0$	$4,8$	$2$
$2$	$6,10$	$2$
$4$	$0,8$ (However, case of $0,4$ is already taken)	$1$
$6$	$2,10$ (However, case of $2,6$ is already taken)	$1$
$8$	$0,4$ (However, both cases have been taken)	$0$
$10$	$2,6$ (However, both cases have been taken)	$0$

Hence, total number of cases possible are $2+2+1+1=6$

Total ways to choose $2$ numbers from $\{0,1,2,...,10\}$ are ${}^{11}{C}_2=55$

Hence, probability $=\frac{6}{55}$