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.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
The pairs (a,b) where both a+b and |a-b| are multiples of 4 require a+b ≡ 0 (mod 4) AND a-b ≡ 0 (mod 4), which together imply 2a ≡ 0 (mod 4) and 2b ≡ 0 (mod 4), so both a and b must be multiples of 2. Further, since a+b ≡ 0 (mod 4) and both are even, writing a=2m, b=2n gives 2(m+n) ≡ 0 (mod 4), so m+n must be even, meaning m and n have the same parity — both odd or both even.
From {0,1,...,10}, the even numbers are {0,2,4,6,8,10}. Among these, even multiples of 2: those where m=a/2 is even → a ∈ {0,4,8} and those where m is odd → a ∈ {2,6,10}.
Valid pairs: choose 2 from {0,4,8} → C(3,2)=3 pairs, or choose 2 from {2,6,10} → C(3,2)=3 pairs. Total = 6 favorable pairs.
Total ways to choose 2 from 11 numbers = C(11,2) = 55. Probability = 6/55.
From {0,1,...,10}, the even numbers are {0,2,4,6,8,10}. Among these, even multiples of 2: those where m=a/2 is even → a ∈ {0,4,8} and those where m is odd → a ∈ {2,6,10}.
Valid pairs: choose 2 from {0,4,8} → C(3,2)=3 pairs, or choose 2 from {2,6,10} → C(3,2)=3 pairs. Total = 6 favorable pairs.
Total ways to choose 2 from 11 numbers = C(11,2) = 55. Probability = 6/55.
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