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 multiples of 4, is

Options

A. `(6)/(55)`
B. `(12)/(55)`
C. `(14)/(45)`
D. `(7)/(55)`

Answer

A

Explanation

There are 11 numbers out of which two numbers a and b can be chosen in `.^(11)C_(2)` ways.
If a+b and a-b both are disable by 4, then there exist `lamda " and " mu` such that `a+b=4lamda " and " a-b=4 mu`.
`therefore 2a=4(lamda+mu),2b=4(lamda-mu) implies a=2 (lamda+ mu),b=2(lamda-mu)`
Various possibilities of a and b are :

So, favorable number of events =6
Hence, required probability `=(6)/(.^(11)C_(2))=(6)/(55)`