If two different numbers are taken from the set {0, 1, 2, 3, …, 10}; then the probability that their

Q: Wait, why do we only consider even numbers? Can't an odd number work?

Great question! Let's look at the algebra. If $x+y = 4k$ and $x-y = 4m$, adding them gives $2x = 4(k+m)$, so $x = 2(k+m)$, which is clearly even. Similarly, subtracting them gives $2y = 4(k-m)$, so $y$ is also even. If either $x$ or $y$ were odd, their sum or difference would be odd, which cannot be a multiple of 4. So, yes, both must be even!

Q: Why does $|x-y| = 4m$ mean $x \equiv y \pmod 4$?

Think of modular arithmetic as a clock. Saying $|x-y|$ is a multiple of 4 means that the distance between $x$ and $y$ on the number line is a multiple of 4. If you start at $x$ and jump by 4s, you land on numbers that have the same remainder as $x$ when divided by 4. Thus, $x$ and $y$ must belong to the same 'remainder class' modulo 4.

Q: Can we pick the same number twice? The problem says 'two different numbers'.

You are absolutely right to be cautious! The problem explicitly states 'two different numbers'. This is why we use the combination formula $^{n}C_{2}$ rather than allowing repetitions. We are selecting a subset of size 2 from our groups, ensuring the numbers are distinct.

Question

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

Options:

(A) $\frac{6}{55}$

(B) $\frac{12}{55}$

(C) $\frac{14}{45}$

(D) $\frac{7}{55}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a probability problem; we are embarking on a journey through the elegant landscape of number theory.

First, let us define our universe. We are working with the set $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, which contains exactly 11 integers.

The problem asks us to select two different numbers from this set. The total number of ways to choose 2 distinct elements out of 11 is given by the combination formula $^{11}C_{2}$.

$$^{11}C_{2} = \frac{11 	imes 10}{2} = 55$$

There are 55 possible pairs in our sample space. This value serves as the denominator of our probability fraction.

### The Parity Revelation

Now, let us look at the constraints. We are looking for pairs $(x, y)$ such that their sum $x+y$ is a multiple of 4, and their absolute difference $|x-y|$ is also a multiple of 4.

Let us write these as $x+y = 4k$ and $|x-y| = 4m$ for some integers $k$ and $m$.

If we consider the case where $x > y$, we have $(x+y) + (x-y) = 4k + 4m$, which simplifies to $2x = 4(k+m)$, or $x = 2(k+m)$. This implies that $x$ must be an even number.

Similarly, subtracting the equations yields $2y = 4(k-m)$, or $y = 2(k-m)$, meaning $y$ must also be an even number. We can now discard all odd numbers from our set, leaving us with the relevant subset: $\{0, 2, 4, 6, 8, 10\}$.

### The Modular Dance

We have established that both numbers must be even. However, we must also satisfy the condition $|x-y| = 4m$, which is equivalent to $x \equiv y \pmod 4$.

This means that when we divide $x$ and $y$ by 4, they must leave the same remainder. Let us categorize our even numbers by their remainders modulo 4:

- $0 \equiv 0 \pmod 4$
- $2 \equiv 2 \pmod 4$
- $4 \equiv 0 \pmod 4$
- $6 \equiv 2 \pmod 4$
- $8 \equiv 0 \pmod 4$
- $10 \equiv 2 \pmod 4$

We have two distinct families of numbers. Group 1 (remainder 0) contains $\{0, 4, 8\}$, and Group 2 (remainder 2) contains $\{2, 6, 10\}$. To satisfy the condition that the difference is a multiple of 4, both numbers must belong to the same group.

### Final Calculation

Now, the counting becomes straightforward. We need to choose two numbers from the same group:

- In Group 1, we have 3 numbers. The number of ways to choose 2 is $^{3}C_{2} = 3$.
- In Group 2, we have 3 numbers. The number of ways to choose 2 is $^{3}C_{2} = 3$.

Adding these together, we get $3 + 3 = 6$ favorable pairs. The probability is the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{6}{55}$$

By peeling back the layers of the problem—first by identifying the parity, then by partitioning the set using modular arithmetic—we have arrived at the final answer of **$\frac{6}{55}$**.

Animated Solution for Mathematics - Probability: If two different numbers are taken from the set ${0, 1, 2, 3, \dots, 10}$ ; then the probability that their sum as well as absolute difference are both multiple of 4, is:

Select Answer:

Visualized Solution

Defining the Sample Space $S$

Calculating Total Outcomes

Analyzing the Conditions

Deducing Parity of $x$ and $y$

The Remainder Constraint

Grouping by Remainder mod 4

Counting Favorable Pairs

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Parity Revelation

The Modular Dance

Final Calculation

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