If two different numbers are taken from the set {0,1,2,3, ......, 10} ... If two different numbers are taken from the set {0,1,2,3, ......, 10}  then what is the probability that their sum as well as absolute difference are both multiples of 4
Here is my work out
The sample space here is equal to 55.
Now to me the possible combinations are {0,4},{0,8},{2,6},{2,10},{4,8},{6,10}
so to me the answer is 6/55
 A: Let's think what it means for two numbers $a, b$ to have sum and difference multiples of 4.
$a+b\equiv0 \pmod4$
$a-b\equiv0 \pmod4$
Adding them up, $2a\equiv0 \pmod4$, so it follows that $a\equiv0,2 \pmod4$
$a-b\equiv0 \pmod4$ states that $a\equiv b \pmod4$.
So, $a, b$ are $0, 2 \pmod4$ and are congruent to each other. This splits nicely into 2 cases, $0 \pmod4$ and $2 \pmod4$.
Under the case $0 \pmod4$, $a, b$ can be $0, 4, 8$. There are 3 ways to select 2 numbers from this set, and it is easy to verify that all 3 ways work.
Under the case $2 \pmod4$, $a, b$ can be $2, 6, 10$. There are 3 ways to select 2 numbers from this set, and it is easy to verify that all 3 ways work.
With $6$ satifying possibilities in total, and ${11\choose2}=55$ ways to choose 2 numbers, the probability of this event is $\frac{6}{55}$.
A: 6/55 is correct sample space I {(0,4),(4,0),(0,8),(8,0),(2,6),(6,2),(2,10),(10,2),(4,8),(8,4),(6,10),(10,6)} terms are 12 
And total number of ways we can arrange them is 11*10 
Which is 110 now  12/110 is 6/55 
A: Assuming random selection-- To have both sum and absolute difference of different integers be a multiple of 4 there are 5 possible combination of integers.0 and 4 , 2 and 6, 4 and 8, 6 and 10, and 0 and 8, = 5 choices
Given that  choice of 2 numbers from 0 to 10 , that is 11 choices, and that they must be different, the total choices is 11x10 = 110.
The probability of the 5 combinations being chosen, remembering that we halve because there is no requirement for order, is 5 (times 2) divided by 110  = 1/11
