In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q60 jee_main_2025_29_jan_evening Rank of a Word in Dictionary

If all the words with or without meaning made using all the letters of the word “KANPUR” are arranged as in a dictionary, then the word at

440th

position in this arrangement, is :

A. $PRNAKU$ $\text{PRNAKU}$
B. $PRKANU$ $\text{PRKANU}$
C. $PRKAUN$ $\text{PRKAUN}$
D. $PRNAUK$ $\text{PRNAUK}$

Solution

Related Formula

Number of permutations of $n$ $n$ distinct letters is given by:

Permutations = n!

Core Logic

Sort the distinct letters of "KANPUR" alphabetically:

A, K, N, P, R, U

Track alphabetical structural sets row by row: Words starting with A: $5! = 120$ $5! = 120$ Words starting with K: $5! = 120$ $5! = 120$ (Cumulative: 240) Words starting with N: $5! = 120$ $5! = 120$ (Cumulative: 360)

Step 1: Parse the Next Character Layer

We need the 440th word, so the next block begins with P: Words starting with PA: $4! = 24$ $4! = 24$ (Cumulative: 384) Words starting with PK: $4! = 24$ $4! = 24$ (Cumulative: 408) Words starting with PN: $4! = 24$ $4! = 24$ (Cumulative: 432)

Step 2: Reach the Targeted Count

Remaining difference to hit 440 is $440 - 432 = 8$ $440 - 432 = 8$ words. The next group starts with PR. Alphabetical listings within PR: Words starting with PRA: $3! = 6$ $3! = 6$ (Cumulative: 438)

Now list alphabetically within PRK: 439th word: PRKANU 440th word: PRKAUN

Pattern Recognition

Factorial blocks eliminate large numbers quickly. Once the cumulative total lands within a narrow range of the answer, switch over to writing combinations down manually to ensure high accuracy.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q jee_main_2025_28_jan_morning Permutations under Restrictions

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is

A. 4608
B. 5720
C. 5719
D. 4607

Solution

Related Formula

For a 5-digit number, total permutations with repetition allowed for $n$ $n$ digits is given by:

Total Cases = d₁ \times d₂ \times d₃ \times d₄ \times d₅

Core Logic

We need 5-digit numbers greater than 50000 using digits $0, 1, 2, 3, 4, 5, 6, 7$ $\{0, 1, 2, 3, 4, 5, 6, 7\}$ under the restriction $d₁ + d₅ \leq 8$ $d_1 + d_5 \le 8$.

Let's analyze the pairs $(d₁, d₅)$ $(d_1, d_5)$ where $d₁ in 5, 6, 7$ $d_1 \in \{5, 6, 7\}$: Case I: $d₁ = 5 \Rightarrow d₅ in 0, 1, 2, 3$ (4 options) Case II: $d₁ = 6 \Rightarrow d₅ in 0, 1, 2$ (3 options) Case III: $d₁ = 7 \Rightarrow d₅ in 0, 1$ $d_1 = 7 \Rightarrow d_5 \in \{0, 1\}$ (2 options)

Total choices for the first and last digits combined = $4 + 3 + 2 = 9$ $4 + 3 + 2 = 9$ pairs.

Step 1: Calculating Intermediate Choices

The middle three digits ( $d₂, d₃, d₄$ $d_2, d_3, d_4$) have no restrictions and can each be chosen from any of the 8 available digits.

Number of ways = 9 \times (8 \times 8 \times 8) = 4608

Step 2: Subtracting Boundary Conditions

Since the question specifies numbers strictly greater than 50000, we must check if 50000 is included in our count. For $d₁=5$ $d_1=5$ and $d₅=0$ $d_5=0$, setting $d₂=d₃=d₄=0$ $d_2=d_3=d_4=0$ gives exactly 50000, which is included in the 4608 count.

Total numbers = 4608 - 1 = 4607

Pattern Recognition

Always look carefully at edge constraints like "greater than". Counting the number 50000 explicitly avoids typical off-by-one errors.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q jee_main_2025_28_jan_morning Combinatorial Coefficients and Locus

Let

ⁿCr - 1 = 28, ⁿCᵣ = 56

and

ⁿCr + 1 = 70

. Let A(4cost, 4sint), B(2sint, -2cost) and C(3r - n,

r² - n - 1

) be the vertices of a triangle ABC, where t is a parameter. If

(3x - 1)² + (3y)² = α

, is the locus of the centroid of triangle ABC, then

α

equals:

A. 20
B. 8
C. 6
D. 18

Solution

Related Formula

Consecutive combinations ratio property:

ⁿCᵣ₋₁ⁿCᵣ = (r)/(n-r+1)

Core Logic

Setting up ratios between consecutive given coefficients:

(28)/(56) = (1)/(2) = (r)/(n-r+1) 3r = n + 1 (1)

(56)/(70) = (4)/(5) = (r+1)/(n-r) 9r = 4n - 5 (2)

Solving equations (1) and (2) gives $r = 3$ $r = 3$ and $n = 8$ $n = 8$.

Step 1: Locating Vertices and Centroid Locus

Substituting values for point $C$ $C$ gives $C(1,0)$ $C(1,0)$. Let the centroid coordinates be $(x,y)$ $(x,y)$:

3x = 4 t + 2 t + 1 3x - 1 = 4 t + 2 t

3y = 4 t - 2 t + 0 3y = 4 t - 2 t

Step 2: Squaring and Summing Trig Components

Squaring and adding both parametric tracking components eliminates $t$ $t$:

(3x - 1)² + (3y)² = (4 t + 2 t)² + (4 t - 2 t)² = 16 + 4 = 20

Thus, $α = 20$ $\alpha = 20$.

Pattern Recognition

Symmetric parameter sets of form $(A t + B t)² + (A t - B t)²$ collapse instantly into $A² + B²$ $A^2 + B^2$ via basic Pythagorean identities.

Chapter Mix

Class 11 Maths: Straight Lines Class 11 Maths: Permutations and Combinations

Q74 jee_main_2025_03_april_morning Numbers and Digits Sum Criteria

If the number of seven-digit numbers, such that the sum of their digits is even, is

m \cdot n \cdot 10ⁿ

, where

m, n in 1, 2, 3, , 9

, then

m + n

is equal to:

Numerical Answer. Answer: 14 to 14

Solution

Related Formula

Parity property for numerical spaces: Across any continuous span sequence ending in zero, exactly half of the configuration combinations form an even sum of digits while the rest form odd outputs.

Core Logic

Calculate the total possible combinations of 7-digit numbers first:

Total Numbers = 9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 9,000,000

By using fundamental parity distributions across digit configurations, exactly half of these total options have an even sum of digits:

Even Sum Count = (9,000,000)/(2) = 4,500,000

Step 1: Extracting factors

Express the final numeric amount in requested base-10 exponential format shape: $4,500,000 = 45 \times 10⁵ = 9 \cdot 5 \cdot 10⁵$ [cite: 1468]

Matching structural parameters: $m = 9, n = 5$ $$m = 9, \quad n = 5$$ [cite: 1469] Both numbers belong to set range $1, 2, , 9$ $\{1, 2, \dots, 9\}$.

Find target summation parameter value [cite: 1470]:

m + n = 9 + 5 = 14

Pattern Recognition

The last slot digit completely decides final parity choices. This allows splitting total permutation blocks directly in half without tedious calculations.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q74 jee_main_2025_04_april_evening Counting Principles

Let

m

and

n

,

(m < n)

be two 2-digit numbers. Then the total numbers of pairs

(m, n)

, such that

(m, n) = 6

, is

Numerical Answer. Answer: 64 to 64

Solution

Core Logic

Since $(m,n) = 6$ $\gcd(m,n) = 6$, we can define $m = 6a$ $m = 6a$ and $n = 6b$ $n = 6b$, where $a$ $a$ and $b$ $b$ are coprime integers ( $(a,b) = 1$ $\gcd(a,b) = 1$). Given that $m < n$ $m < n$, we must have $a < b$ $a < b$. Both $m$ $m$ and $n$ $n$ are two-digit numbers, which means $10 \leq m, n \leq 99$ $10 \le m, n \le 99$:

10 \leq 6a \leq 99 1.66 \leq a \leq 16.5 2 \leq a \leq 16

10 \leq 6b \leq 99 1.66 \leq b \leq 16.5 2 \leq b \leq 16

Thus, we need to count all coordinate integer pairs $(a,b)$ $(a,b)$ satisfying $2 \leq a < b \leq 16$ $2 \le a < b \le 16$ such that $(a,b) = 1$ $\gcd(a,b) = 1$.

Step 1: Systematic Counting by Fixed Value of 'a'

Let's list the valid values for $b$ $b$ for each choice of $a$ $a$ in the range $[2, 16]$ $[2, 16]$:

$a=2$ $a=2$: $b in 3, 5, 7, 9, 11, 13, 15 7 pairs$
$a=3$ $a=3$: $b in 4, 5, 7, 8, 10, 11, 13, 14, 16 9 pairs$
$a=4$ $a=4$: $b in 5, 7, 9, 11, 13, 15 6 pairs$
$a=5$ $a=5$: $b in 6, 7, 8, 9, 11, 12, 13, 14, 16 9 pairs$
$a=6$ $a=6$: $b in 7, 11, 13 3 pairs$
$a=7$ $a=7$: $b in 8, 9, 10, 11, 12, 13, 15, 16 8 pairs$
$a=8$ $a=8$: $b in 9, 11, 13, 15 4 pairs$
$a=9$ $a=9$: $b in 10, 11, 13, 14, 16 5 pairs$
$a=10$ $a=10$: $b in 11, 13 2 pairs$
$a=11$ $a=11$: $b in 12, 13, 14, 15, 16 5 pairs$
$a=12$ $a=12$: $b in 13 1 pair$ $b \in \{13\} \implies 1 \text{ pair}$
$a=13$ $a=13$: $b in 14, 15, 16 3 pairs$
$a=14$ $a=14$: $b in 15 1 pair$ $b \in \{15\} \implies 1 \text{ pair}$
$a=15$ $a=15$: $b in 16 1 pair$ $b \in \{16\} \implies 1 \text{ pair}$

Step 2: Final Summation

Summing up all valid ordered coordinate tracking entries:

Total = 7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1 + 3 + 1 + 1 = 64

Pattern Recognition

For modular subset counts, convert your boundary targets to factor conditions directly. Listing terms by prime factors reduces counting errors compared to checking every pair from scratch.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Number Theory

Solution & Explanation

Related Formula

Core Logic

Step 1: Identifying Valid Selection Cases

Step 2: Calculating Combinations per Case

Step 3: Summing the Total Ways

Pattern Recognition

Chapter Mix

Reference Study Guides

More Permutations and Combinations Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Parse the Next Character Layer

Step 2: Reach the Targeted Count

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Intermediate Choices

Step 2: Subtracting Boundary Conditions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Locating Vertices and Centroid Locus

Step 2: Squaring and Summing Trig Components

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extracting factors

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Systematic Counting by Fixed Value of 'a'

Step 2: Final Summation

Pattern Recognition

Chapter Mix

More Permutations and Combinations Questions — jee_main_2024_30_january_evening

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