Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ..., 20}. Statement-1

Q: Why do we use combinations (${}^{20}C_4$) instead of permutations?

Great question! We use combinations because we are choosing a *set* of four numbers. The order in which we pick them doesn't matter; what matters is the final collection. Once the set is chosen, if they form an AP, there is only one way to arrange them in increasing order. The set itself is the fundamental unit of our sample space.

Q: I thought $d$ could only go up to 5. Why is $d=6$ valid?

That is the classic trap! If $d=6$, the sequence is $a, a+6, a+12, a+18$. If $a=1$, the sequence is $1, 7, 13, 19$, which is $\le 20$. It fits! Always trust the inequality $a+3d \le 20$ over your intuition. Never assume the bounds without checking the math.

Q: How did you get the formula $20-3d$?

Think of it as a sliding window. You have 20 slots. An AP of difference $d$ occupies a span of $3d$. You are sliding this 'rigid rod' of length $3d$ across the 20 slots. The number of positions is $20 - (3d + 1) + 1 = 20 - 3d$. It is a beautiful geometric way to count!

Question

Four numbers are chosen at random (without replacement) from the set $\{1, 2, 3, ..., 20\}$. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is $1/85$. Statement-2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $\{\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\}$.

Options:

(A) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is true, Statement-2 is false

(C) Statement-1 is false, Statement-2 is true

(D) Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1

Sigma Percentile · Accepted Answer

### The Universe of Possibilities

Every probability problem begins with the sample space. We have 20 distinct numbers, and we need to choose 4. Since the order of selection does not change the set, we use combinations.

The total number of ways to choose these numbers is given by ${}^{20}C_4$. Calculating this, we get:

$$n(S) = \frac{20 	imes 19 	imes 18 	imes 17}{4 	imes 3 	imes 2 	imes 1} = 4845$$

This is our universe. Every possible outcome lives within this number.

### The Geometry of the AP

Now, let us define what it means to be an Arithmetic Progression (AP). If we have four numbers in an AP, they must look like $a, a+d, a+2d, a+3d$.

Here, $a$ is the first term and $d$ is the common difference. The crucial constraint is that the largest term, $a+3d$, cannot exceed 20.

So, we have the inequality:

$$a + 3d \le 20$$

Since $a \ge 1$, we can rewrite this as $1 + 3d \le 20$, which simplifies to $3d \le 19$, or $d \le 6.33$. This is where many students stumble; $d$ can be as large as 6.

### The Hunt for the Common Difference

Let us test our bounds. The number of APs for any fixed $d$ is given by $20 - 3d$. We sum these up for all valid $d$ from 1 to 6:

$$n(E) = (20-3) + (20-6) + (20-9) + (20-12) + (20-15) + (20-18)$$

Calculating the sum:

$$n(E) = 17 + 14 + 11 + 8 + 5 + 2 = 57$$

For instance, if $d=6$, we have $20 - 3(6) = 2$ possible APs. Specifically, the sets are $\{1, 7, 13, 19\}$ and $\{2, 8, 14, 20\}$.

### The Final Calculation

Now, the probability is simply the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{57}{4845}$$

Dividing both the numerator and the denominator by 57, we arrive at the elegant result:

$$P(E) = \frac{1}{85}$$

**Statement 1 is true.** Statement 2, however, claims $d$ is limited to $\{\pm 1, \dots, \pm 5\}$.

We have just proven that $d=6$ is a valid common difference. Therefore, **Statement 2 is false.**

Select Answer:

Visualized Solution

Total Sample Space $n (S)$

Defining the AP Condition

Finding Range of $d$

Counting Favorable APs ( $d = 1$ )

Counting Favorable APs ( $d = 6$ )

Total Favorable Outcomes $n (E)$

Verifying Statement-1

Evaluating Statement-2

The Sigma Insight: Classical Definition of Probability

The Universe of Possibilities

The Geometry of the AP

The Hunt for the Common Difference

The Final Calculation

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Select Answer:

Visualized Solution

Total Sample Space .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(S)

Defining the AP Condition

Finding Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Counting Favorable APs ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=1)

Counting Favorable APs ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=6)

Total Favorable Outcomes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n(E)

Verifying Statement-1

Evaluating Statement-2

The Sigma Insight: Classical Definition of Probability

The Universe of Possibilities

The Geometry of the AP

The Hunt for the Common Difference

The Final Calculation

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