JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Probability of Specific Ranks in an Ordered Selection

Q: Why do we not multiply by 5! for arranging the numbers?

Because the problem specifies they are arranged in increasing order. For any 5 chosen distinct numbers, there is exactly one way to arrange them in increasing order. Both the total outcomes and favorable outcomes are sets of 5 elements, so the arrangement factor is 1 in both numerator and denominator.

Five numbers x_1, x_2, x_3, x_4, x_5 are randomly selected from the numbers 1,2,3, … …, 18 and are arranged in the increasing order (x_1

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What feels right?

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A1/136B1/72C1/68D1/34

Step-by-step solutionView

Correct answer

The probability that $x_{2} = 7$ and $x_{4} = 11$ when five distinct numbers are chosen from $1, \dots, 18$ and sorted is $\frac{1}{68}$ .

Option analysis

Why each option works or fails

A · 1/136

You may have doubled the denominator or incorrectly evaluated the combinations by a factor of 2. Compute the ratio directly as $\frac{(\binom{6}{1}) \times (\binom{3}{1}) \times (\binom{7}{1})}{(\binom{18}{5})} = \frac{6 \times 3 \times 7}{8568} = \frac{126}{8568} = \frac{1}{68}$ .

B · 1/72

You miscounted the available integers between 7 and 11 or above 11, using boundary counts with an off-by-one error. The integers strictly between 7 and 11 are $8, 9, 10$ , giving exactly 3 choices for $x_{3}$ , and strictly greater than 11 are $12, \dots, 18$ , giving exactly 7 choices for $x_{5}$ .

C · 1/68

This is the correct calculation. Correctly identifying that $x_{1} \in {1, \dots, 6}$ , $x_{3} \in {8, 9, 10}$ , and $x_{5} \in {12, \dots, 18}$ gives $6 \times 3 \times 7 = 126$ favorable subsets out of $(\binom{18}{5}) = 8568$ , which simplifies to $\frac{1}{68}$ .

D · 1/34

You likely cancelled an extra factor of 2 in the denominator during fraction reduction. Ensure arithmetic in simplifying $\frac{126}{8568}$ is exact: $8568 / 126 = 68$ , so the value is $\frac{1}{68}$ , not $\frac{1}{34}$ .

Reviewed route

Solution

StepWorking

01Given

Five distinct numbers $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}$ are chosen from ${1, 2, 3, \dots, 18}$ . We are given conditions $x_{2} = 7$ and $x_{4} = 11$ .

02Goal

Compute the probability $P (E) = \frac{n (E)}{n (S)}$ where $E$ is the event ${x_{2} = 7 and x_{4} = 11}$ .

03Approach

Any set of 5 distinct numbers from 1 to 18 has a unique strictly increasing arrangement, so total outcomes is $n (S) = (\binom{18}{5})$ . For favorable outcomes, fixing $x_{2} = 7$ and $x_{4} = 11$ requires selecting $x_{1} \in {1, \dots, 6}$ , $x_{3} \in {8, 9, 10}$ , and $x_{5} \in {12, \dots, 18}$ . Then apply the multiplication rule.

04Execute

The number of choices are: - For $x_{1}$ : chosen from ${1, 2, 3, 4, 5, 6} ⟹ (\binom{6}{1}) = 6$ ways. - For $x_{3}$ : chosen from ${8, 9, 10} ⟹ (\binom{3}{1}) = 3$ ways. - For $x_{5}$ : chosen from ${12, 13, 14, 15, 16, 17, 18} ⟹ (\binom{7}{1}) = 7$ ways. Thus, $n (E) = 6 \times 3 \times 7 = 126$ . The total sample space size is: $n (S) = (\binom{18}{5}) = \frac{18 \times 17 \times 16 \times 15 \times 14}{5 \times 4 \times 3 \times 2 \times 1} = 8568$ . Therefore, $P (E) = \frac{126}{8568} = \frac{126}{126 \times 68} = \frac{1}{68}$ .

✓Verify

Check arithmetic: $n (S) = 18 \times 17 \times 2 \times 14 = 8568$ . Dividing by $n (E) = 126$ : $8568 / 126 = 68$ . Thus $P (E) = 1 / 68$ .

Hints that build this answer step by step

Given $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}$ with $x_{2} = 7$ and $x_{4} = 11$ , from which sets must $x_{1}$ , $x_{3}$ , and $x_{5}$ be chosen?
$x_{1} \in {1, \dots, 6}$ , $x_{3} \in {8, 9, 10}$ , and $x_{5} \in {12, \dots, 18}$
How many favorable selections of $(x_{1}, x_{3}, x_{5})$ exist?
$6 \times 3 \times 7 = 126$
What is the total number of ways to choose 5 distinct numbers from ${1, 2, \dots, 18}$ , and what is the resulting probability?
Total is $(\binom{18}{5}) = 8568$ , giving a probability of $\frac{126}{8568} = \frac{1}{68}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why do we not multiply by 5! for arranging the numbers?

Because the problem specifies they are arranged in increasing order. For any 5 chosen distinct numbers, there is exactly one way to arrange them in increasing order. Both the total outcomes and favorable outcomes are sets of 5 elements, so the arrangement factor is 1 in both numerator and denominator.

Answer

The probability that $x_{2} = 7$ and $x_{4} = 11$ when five distinct numbers are chosen from $1, \dots, 18$ and sorted is $\frac{1}{68}$ .

Why each option works or fails

A: 1/136 - You may have doubled the denominator or incorrectly evaluated the combinations by a factor of 2. Compute the ratio directly as $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}$ .
B: 1/72 - You miscounted the available integers between 7 and 11 or above 11, using boundary counts with an off-by-one error. The integers strictly between 7 and 11 are $8, 9, 10$ , giving exactly 3 choices for $x_{3}$ , and strictly greater than 11 are $12, \dots, 18$ , giving exactly 7 choices for $x_{5}$ .
C · correct: 1/68 - This is the correct calculation. Correctly identifying that $x_{1} \in {1, \dots, 6}$ , $x_{3} \in {8, 9, 10}$ , and $x_{5} \in {12, \dots, 18}$ gives $6 \times 3 \times 7 = 126$ favorable subsets out of $(\binom{18}{5}) = 8568$ , which simplifies to $\frac{1}{68}$ .
D: 1/34 - You likely cancelled an extra factor of 2 in the denominator during fraction reduction. Ensure arithmetic in simplifying $\frac{126}{8568}$ is exact: $8568 / 126 = 68$ , so the value is $\frac{1}{68}$ , not $\frac{1}{34}$ .

Step-by-step solution

Given: Five distinct numbers $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}$ are chosen from ${1, 2, 3, \dots, 18}$ . We are given conditions $x_{2} = 7$ and $x_{4} = 11$ .
Goal: Compute the probability $P (E) = \frac{n (E)}{n (S)}$ where $E$ is the event ${x_{2} = 7 and x_{4} = 11}$ .
Approach: Any set of 5 distinct numbers from 1 to 18 has a unique strictly increasing arrangement, so total outcomes is $n (S) = (\binom{18}{5})$ . For favorable outcomes, fixing $x_{2} = 7$ and $x_{4} = 11$ requires selecting $x_{1} \in {1, \dots, 6}$ , $x_{3} \in {8, 9, 10}$ , and $x_{5} \in {12, \dots, 18}$ . Then apply the multiplication rule.
Execute: The number of choices are: - For $x_{1}$ : chosen from ${1, 2, 3, 4, 5, 6} ⟹ (\binom{6}{1}) = 6$ ways. - For $x_{3}$ : chosen from ${8, 9, 10} ⟹ (\binom{3}{1}) = 3$ ways. - For $x_{5}$ : chosen from ${12, 13, 14, 15, 16, 17, 18} ⟹ (\binom{7}{1}) = 7$ ways. Thus, $n (E) = 6 \times 3 \times 7 = 126$ . The total sample space size is: $n (S) = (\binom{18}{5}) = \frac{18 \times 17 \times 16 \times 15 \times 14}{5 \times 4 \times 3 \times 2 \times 1} = 8568$ . Therefore, $P (E) = \frac{126}{8568} = \frac{126}{126 \times 68} = \frac{1}{68}$ .
Verify: Check arithmetic: $n (S) = 18 \times 17 \times 2 \times 14 = 8568$ . Dividing by $n (E) = 126$ : $8568 / 126 = 68$ . Thus $P (E) = 1 / 68$ .

Probability of Specific Ranks in an Ordered Selection

What feels right?

The probability that $x_{2} = 7$ and $x_{4} = 11$ when five distinct numbers are chosen from $1, \dots, 18$ and sorted is $\frac{1}{68}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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