Let |X| denote the number of elements in set X. Let S = {1,2,3,4,5,6} be a sample space, where each

Q: Wait, why does $c = \frac{ab}{6}$ have to be an integer? Isn't it just a probability equation?

That is a brilliant observation! While the equation starts as a probability relation, $P(A \cap B) = \frac{|A \cap B|}{6}$. Since $|A \cap B|$ represents the count of elements in the intersection, it must be a non-negative integer. This constraint is the 'hidden key' that unlocks the entire problem.

Q: Why did we treat the case $a=6$ differently from the others?

When $a=6$, the set $A$ is the entire sample space $S$. If $A=S$, then $A \cap B = B$. The independence condition $P(A \cap B) = P(A)P(B)$ becomes $P(B) = 1 \cdot P(B)$, which is always true for any subset $B$. Thus, we don't need to worry about the divisibility constraint for $a=6$; we just need to count all possible non-empty, proper subsets $B$.

Q: I'm confused about the combinatorics step. Why are we using $\binom{6}{c} \times \binom{6-c}{a-c} \times \binom{6-a}{b-c}$?

Think of the sample space as a collection of $6$ distinct slots. We are partitioning these slots into four disjoint regions: $(A \cap B)$, $(A \setminus B)$, $(B \setminus A)$, and the region outside both. The formula you see is just a systematic way of filling these 'buckets' based on the sizes $a, b,$ and $c$ we derived.

Q: Can $b$ be equal to $a$? The problem says $1 \leq |B| < |A|$.

You are absolutely right to check that! The strict inequality $|B| < |A|$ is a crucial constraint. If $b$ were equal to $a$, we would have to include those cases, but the problem explicitly excludes them. Always double-check the inequality signs in JEE problems—they are often where the 'trap' lies.

Q: Is there any other case where $ab$ is a multiple of $6$ that we missed?

We checked all possibilities for $a \in \{2, 3, 4, 5, 6\}$. For $a=5$, $5b$ is a multiple of $6$ only if $b$ is a multiple of $6$, which forces $b=6$, violating $b < a$. For $a=2$, $2b$ is a multiple of $6$ only if $b$ is a multiple of $3$, which forces $b=3$, violating $b < a$. We have been exhaustive!

Question

Let $|X|$ denote the number of elements in set $X$. Let $S = \{1,2,3,4,5,6\}$ be a sample space, where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$, then the number of ordered pairs $(A,B)$ such that $1 \leq |B| < |A|$, equals

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a sample space $S$ containing $6$ distinct, equally likely elements. You are tasked with choosing two subsets, $A$ and $B$, such that they are statistically independent.

In the world of probability, independence is a powerful, elegant symmetry. It tells us that the occurrence of $A$ provides absolutely no information about the occurrence of $B$.

Mathematically, this is defined by the beautiful relation $P(A \cap B) = P(A)P(B)$. But how does this abstract concept translate into the concrete reality of counting sets?

### The Hidden Constraint

First, let us define our variables. Let $|A| = a$ and $|B| = b$. Since our sample space has $6$ elements, the probability of any event $E$ is simply $\frac{|E|}{6}$.

Substituting this into our independence condition, we get:

$$\frac{|A \cap B|}{6} = \frac{a}{6} 	imes \frac{b}{6}$$

If we let $c = |A \cap B|$, the equation simplifies to $c = \frac{ab}{6}$. Here is the moment of truth: $c$ must be an integer because it represents the number of elements in the intersection.

This means $ab$ must be a multiple of $6$. This is not just an algebraic requirement; it is the geometric heartbeat of the problem. It restricts the possible sizes of our sets $A$ and $B$ in a way that feels almost like a puzzle waiting to be solved.

### Exploring the Possibilities

We know that $1 \leq b < a \leq 6$. Let us test the values of $a$ systematically.

If $a=6$, then $ab = 6b$, which is always a multiple of $6$. Since $b < a$, $b$ can be any integer from $1$ to $5$. This is a special case because $A$ is the entire sample space $S$.

If $a=5$, then $5b$ must be a multiple of $6$. This forces $b$ to be a multiple of $6$. But since $b < 5$, there are no solutions here.

If $a=4$, then $4b$ must be a multiple of $6$, which means $2b$ must be a multiple of $3$. Thus, $b$ must be a multiple of $3$. Given $b < 4$, the only solution is $b=3$. This gives us an intersection size $c = \frac{4 	imes 3}{6} = 2$.

If $a=3$, then $3b$ must be a multiple of $6$, meaning $b$ must be even. Given $b < 3$, the only solution is $b=2$. This gives us an intersection size $c = \frac{3 	imes 2}{6} = 1$.

### The Combinatorial Architecture

Now that we have our valid triplets $(a, b, c)$, we must count the number of ways to actually construct these sets. We use the multinomial coefficient logic.

We have $6$ elements to distribute into four distinct regions: the intersection, the part of $A$ only, the part of $B$ only, and the part outside both. The number of ways to choose these is given by:

$$	ext{Ways} = \binom{6}{c} 	imes \binom{6-c}{a-c} 	imes \binom{6-a}{b-c}$$

For the case $a=3, b=2, c=1$, we calculate:

$$\binom{6}{1} 	imes \binom{5}{2} 	imes \binom{3}{1} = 6 	imes 10 	imes 3 = 180$$

For the case $a=4, b=3, c=2$, we calculate:

$$\binom{6}{2} 	imes \binom{4}{2} 	imes \binom{2}{1} = 15 	imes 6 	imes 2 = 180$$

For the case $a=6$, $A$ is fixed as the set $S$. Since $1 \leq |B| < |A|$, $B$ can be any subset of $S$ except the empty set and the set $S$ itself. The number of such subsets is $2^6 - 2 = 64 - 2 = 62$.

### The Grand Total

Summing these up, we find the total number of valid pairs $(A, B)$ is:

$$180 + 180 + 62 = 422$$

We have navigated the constraints, respected the laws of probability, and arrived at the solution. Remember, in JEE Advanced, the math is never just about the final number; it is about the elegance of the path taken.

You have successfully translated a condition of independence into a rigorous counting problem. The final answer is **422**.

Enter Numerical Value:

Visualized Solution

$Sample Space & Events$

$C o n d i t i o n f or I n d e p e n d e n ce$

$R a w S e t u p (S u b s t i t u t i o n)$

$I n t er sec t i o n S i z e R e l a t i o n$

$T h e S i z e C o n s t r ain t$

$C a se 1 : a = 6$

$C h ec k in g a = 5 an d a = 4$

$C h ec k in g a = 3 an d a = 2$

$C o mbina t or i cs S e t u p$

$P ai r s f or a = 3, b = 2, c = 1$

$P ai r s f or a = 4, b = 3, c = 2$

$P ai r s f or a = 6$

$T o t a l O r d er e d P ai r s$

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Hidden Constraint

Exploring the Possibilities

The Combinatorial Architecture

The Grand Total

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Enter Numerical Value:

Visualized Solution

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