A board has 16 squares as shown in the figure:
Out of these 16 squares, two squares are chosen at ra

Q: Wait, why do we use $^{16}C_2$ instead of $16 \times 15$?

Great question! We use $^{16}C_2$ because the order in which we pick the two squares does not matter. If we pick square A then square B, it is the same pair as picking square B then square A. Using combinations ($^{n}C_r$) ensures we count each unique pair exactly once.

Q: How do we know there are exactly 12 horizontal pairs?

Think of a single row of 4 squares: $[1, 2, 3, 4]$. The adjacent pairs are $(1,2)$, $(2,3)$, and $(3,4)$. That is 3 pairs per row. Since there are 4 rows, we have $4 \times 3 = 12$ horizontal pairs. The same logic applies to the 4 columns, giving us another 12 vertical pairs.

Q: Is there a general formula for an $n \times n$ grid?

Yes, there is! For an $n \times n$ grid, the number of horizontal adjacent pairs is $n(n-1)$ and the number of vertical adjacent pairs is also $n(n-1)$. So, the total number of adjacent pairs is $2n(n-1)$. For our $4 \times 4$ grid, this is $2 \times 4 \times 3 = 24$, which matches our calculation perfectly.

Question

A board has 16 squares as shown in the figure:
Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:

Options:

(A) 7/10

(B) 4/5

(C) 23/30

(D) 3/5

Sigma Percentile · Accepted Answer

### The Geometry of Choice: A Probability Adventure

Welcome, fellow explorer of mathematics! Today, we are going to unravel a beautiful problem that sits at the intersection of combinatorics and probability.

Imagine a $4 	imes 4$ grid—a classic, sixteen-square board. Our mission is simple to state but requires a sharp mind to execute: if we randomly select two squares from this board, what is the probability that they do not share a side?

### Phase 1: The Landscape of Possibilities

Before we dive into the 'no side' condition, we must understand the total universe of our experiment. We have $16$ squares, and we are choosing $2$.

Since the order of selection does not change the pair, we use the combination formula $^{n}C_{r}$. Our sample space, $n(S)$, is given by $^{16}C_{2}$.

Calculating this, we get:

$$n(S) = \frac{16 	imes 15}{2 	imes 1} = 120$$

So, there are exactly $120$ unique ways to pick two squares from this board. This is our denominator.

### Phase 2: The Power of the Complement

Now, here is where the intuition of a master problem-solver kicks in. If we tried to count all the pairs that *do not* share a side, we would be here all day, drowning in a sea of cases.

Instead, we use the elegant strategy of complementary counting. We ask: what is the probability that the two squares *do* share a side?

Let this be event $A$. The probability we want is simply $P(A') = 1 - P(A)$.

### Phase 3: Counting the Neighbors

To find $P(A)$, we need to count how many pairs of squares share a side. A pair shares a side if they are adjacent horizontally or vertically.

Let's look at the horizontal pairs first. In any single row of $4$ squares, there are $3$ adjacent pairs: $(1,2)$, $(2,3)$, and $(3,4)$. Since there are $4$ rows, we have $4 	imes 3 = 12$ horizontal pairs.

Now, look at the columns. By symmetry, each column of $4$ squares also has $3$ adjacent pairs. With $4$ columns, we have $4 	imes 3 = 12$ vertical pairs.

Adding these together, the total number of favorable outcomes for event $A$ is:

$$n(A) = 12 + 12 = 24$$

### Phase 4: The Final Synthesis

We are almost there! The probability of choosing two squares that share a side is:

$$P(A) = \frac{n(A)}{n(S)} = \frac{24}{120}$$

Simplifying this fraction, we divide both numerator and denominator by $24$, yielding $\frac{1}{5}$.

Finally, we apply our complementary rule:

$$P(A') = 1 - \frac{1}{5} = \frac{4}{5}$$

### The Takeaway

Isn't it satisfying when a complex problem collapses into such a clean result? We found that there is an **$80\%$ chance** that two randomly chosen squares will be strangers, sharing no common boundary.

Remember this: whenever you face a counting problem that feels overwhelming, look for the complement. It is often the shortest path to the truth. Keep this logic in your toolkit, and you will be ready for any challenge the JEE throws your way!

Animated Solution for Mathematics - Probability: A board has 16 squares as shown in the figure: Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:

Select Answer:

Visualized Solution

Visualize the $4 \times 4$ Grid

Total Ways to Choose Squares

Calculate Sample Space $n (S)$

The Complementary Strategy

Count Horizontal Adjacent Pairs

Count Vertical Adjacent Pairs

Total Favorable Outcomes $n (A)$

Probability of Sharing a Side $P (A)$

Final Probability $P (A^{'})$

The Way Forward

The Sigma Insight: Classical Definition of Probability

The Geometry of Choice

Phase 1

Phase 2

Phase 3

Phase 4

The Takeaway

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