Probability
Counting with Constraints
Grade 12
Question:
<p>If two different numbers are taken from the set \(\{0, 1, 2, 3, \ldots, 10\}\), then the probability that their sum as well as absolute difference are both multiple of 4, is [JEE Main 2017, 4M]</p>
<p>(a) \(\frac{7}{55}\)</p>
<p>(b) \(\frac{6}{55}\)</p>
<p>(c) \(\frac{12}{55}\)</p>
<p>(d) \(\frac{14}{45}\)</p>
Step-by-Step Solution
Key Concept: Translate the conditions on sum and difference into congruence conditions modulo 4 to identify valid pairs.
<p>Total ways to select 2 different numbers from 11 elements: $\binom{11}{2} = 55$. For sum and absolute difference both divisible by 4, if $a$ and $b$ are the numbers with $a > b$, then $a + b \equiv 0 \pmod{4}$ and $a - b \equiv 0 \pmod{4}$. This implies $2a \equiv 0 \pmod{4}$, so $a \equiv 0 \pmod{2}$ and $2b \equiv 0 \pmod{4}$, so $b \equiv 0 \pmod{2}$. Both must be even with $a \equiv b \pmod{4}$. Favorable pairs: (0,4), (0,8), (4,8), (2,6), (2,10), (6,10) gives 6 pairs.</p>
Correct Answer: B