The number of strictly increasing functions f from the set 1, 2, 3, 4, 5, 6 to the set 1, 2, 3, ldots, 9 such that f(i)…

Year	2026	2025	2024	Total
Questions	13	19	8	40

More Permutations and Combinations Previous-Year Questions

Q23 jee_main_2026_21_jan_morning Divisibility and Counting Rules

Let

S = (m, n): m, n in 1, 2, 3, , 50

. If the number of elements

(m, n)

S

such that

6^m + 9ⁿ

is a multiple of 5 is

p

and the number of elements

(m, n)

S

such that

m + n

is a square of a prime number is

q

, then

p + q

is equal to......

Numerical Answer. Answer: 1333 to 1333

Solution

Related Formula

Modular arithmetic reductions for power cycles:

a \equiv b m \Rightarrow a^k \equiv b^k m

Core Logic

Analyze condition $p$ $p$: $(6^m + 9ⁿ)$ $(6^m + 9^n)$ is divisible by 5. $6 \equiv 1 5 \Rightarrow 6^m \equiv 1^m \equiv 1 5$ . $9 \equiv -1 5 \Rightarrow 9ⁿ \equiv (-1)ⁿ 5$ . For the sum to be divisible by 5: $1 + (-1)ⁿ \equiv 0 5 \Rightarrow (-1)ⁿ = -1$ . This implies $n$ $n$ must be an ODD integer.

Since $m in 1, 2, , 50$ $m \in \{1, 2, \dots, 50\}$, $m$ $m$ can be anything (50 choices). Since $n$ $n$ must be odd in $1, , 50$ $\{1, \dots, 50\}$, $n$ $n$ has 25 choices. $p = 50 \times 25 = 1250$ $p = 50 \times 25 = 1250$.

Step 1: Compute q

Analyze condition $q$ $q$: $(m + n)$ $(m + n)$ is the square of a prime number. Max value of $m+n = 50+50 = 100$ $m+n = 50+50 = 100$. Primes whose squares are $\leq 100$ $\leq 100$: $2, 3, 5, 7$ $2, 3, 5, 7$. Their squares are $4, 9, 25, 49$ $4, 9, 25, 49$. So $m+n$ $m+n$ can be $4, 9, 25, 49$ $4, 9, 25, 49$.

Match List-I with List-II:

	$m+n=4$ $m+n=4$	$m+n=9$ $m+n=9$	$m+n=25$ $m+n=25$	$m+n=49$ $m+n=49$
No. of ways	3	8	24	48

Explanation for counts: If $m+n = S$ $m+n = S$, and $m, n \geq 1$ $m, n \geq 1$, the number of ways is $S-1$ $S-1$ (since $S \leq 50$ $S \leq 50$). For $S=4$ $S=4$: 3 ways. For $S=9$ $S=9$: 8 ways. For $S=25$ $S=25$: 24 ways. For $S=49$ $S=49$: 48 ways. $q = 3 + 8 + 24 + 48 = 83$ $q = 3 + 8 + 24 + 48 = 83$.

Step 2: Final Sum

p + q = 1250 + 83 = 1333

Pattern Recognition

Modular exponentiation immediately shrinks large powers to $\pm 1$ $\pm 1$. The sum $m+n=S$ $m+n=S$ where $1 \leq m,n \leq N$ $1 \le m,n \le N$ has exactly $S-1$ $S-1$ solutions if $S \leq N$ $S \le N$, allowing instant combinatorics tallying without manual counting.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q20 jee_main_2026_21_jan_evening Exponent of Prime in n!

The largest

n in N

, for which

7ⁿ

divides 101!, is:

A. $16$ $16$
B. $18$ $18$
C. $15$ $15$
D. $19$ $19$

Solution

Related Formula

Legendre's Formula: The exponent of a prime p in N! is given by:

Eₚ(N!) = (N)/(p) + (N)/(p²) + (N)/(p³) + \dots

Core Logic

To find the maximum power $n$ $n$ such that $7ⁿ$ $7^n$ divides $101!$ $101!$, we need to find the exponent of the prime $7$ $7$ in the prime factorization of $101!$ $101!$.

Step 1: Apply Legendre's Formula

n = (101)/(7) + (101)/(7²) + (101)/(7³) + \dots

n = (101)/(7) + (101)/(49) + (101)/(343)

$n = 14 + 2 + 0$ $n = 14 + 2 + 0$ $n = 16$ $n = 16$

Pattern Recognition

For prime $p$ $p$ in $N!$ $N!$, iteratively divide $N$ $N$ by $p$ $p$ taking only integer parts and sum them. Fast mental math: $101 \div 7 = 14$ $101 \div 7 = 14$, $14 \div 7 = 2$ $14 \div 7 = 2$. $14+2=16$ $14+2=16$.

Chapter Mix

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

Q25 jee_main_2026_22_january_morning Geometry Based Combinatorics

Let ABC be a triangle. Consider four points

p₁

p₂

p₃

p₄

on the side AB, five points

p₅

p₆

p₇

p₈

p₉

on the side BC and four points

p₁₀

p₁₁

p₁₂

p₁₃

on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points

p₁

p₂

, ....

p₁₃

, is ____.

Numerical Answer. Answer: 660 to 660

Solution

Related Formula

Combinations ⁿCᵣ = (n!)/(r!(n-r)!)

Core Logic

A pentagon requires 5 distinct points as vertices. No three points can be collinear to form a proper polygon. Since the available points lie on the three sides of a triangle, picking 3 points from the same side will form a degenerate straight line segment rather than building a strictly convex vertex frame.

Therefore, we must select the 5 points distributed across the three sides (AB with 4 points, BC with 5 points, AC with 4 points) such that a maximum of 2 points is selected from any one side.

Step 1: Case Breakdown

We need to choose 5 points total from the three groups (4, 5, 4) with the condition that no group contributes more than 2 points. The only valid numerical partitions of 5 into 3 parts bounded by 2 are:

Case 1: 2 points from AB, 2 points from BC, 1 point from AC
Case 2: 2 points from AB, 1 point from BC, 2 points from AC
Case 3: 1 point from AB, 2 points from BC, 2 points from AC

Step 2: Calculating Combinations for Each Case

Case 1: (2 from AB, 2 from BC, 1 from AC)

⁴C₂ \times ⁵C₂ \times ⁴C₁ = 6 \times 10 \times 4 = 240

Case 2: (2 from AB, 1 from BC, 2 from AC)

⁴C₂ \times ⁵C₁ \times ⁴C₂ = 6 \times 5 \times 6 = 180

Case 3: (1 from AB, 2 from BC, 2 from AC)

⁴C₁ \times ⁵C₂ \times ⁴C₂ = 4 \times 10 \times 6 = 240

Step 3: Total Pentagons

Total number of pentagons is the sum of all valid cases:

Total = 240 + 180 + 240 = 660

Geometry Based Combinatorics diagram for Q25 - JEE Main 2026 Morning

Pattern Recognition

For polygon formation from collinear sets, always frame it as a restricted partition problem. A polygon of $k$ $k$ sides requires selecting $k$ $k$ vertices such that no maximum allowable threshold of collinearity is breached (for a strict polygon, no 3 points on a line).

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q25 jee_main_2026_22_january_evening Subsets with Even Product

Let

S

be the set of the first 11 natural numbers. Then the number of elements in

A = B S : n(B) \geq 2 and the product of all elements of B is even

is ____.

Numerical Answer. Answer: 1979 to 1979

Solution

Related Formula

Complementary counting principle:

Favorable Subsets = Total Subsets - Subsets with odd product - Singletons - Empty set

Core Logic

Set $S = 1, 2, 3, , 11$ $S = \{1, 2, 3, \dots, 11\}$ contains 11 elements (6 odd: $1, 3, 5, 7, 9, 11$ $\{1, 3, 5, 7, 9, 11\}$ and 5 even: $2, 4, 6, 8, 10$ $\{2, 4, 6, 8, 10\}$).

Total possible subsets of $S = 2¹¹ = 2048$ $S = 2^{11} = 2048$.
Subsets where product is odd consist entirely of odd numbers: $2⁶ = 64$ $2^6 = 64$.
Singletons with even product: $5$ $5$ (the even numbers themselves).

Step 1: Complementary Subtraction

Excluded subsets:

Empty set (size 0): $1$ $1$
Singletons with odd elements: $6$ $6$
Singletons with even elements: $5$ $5$
Subsets of size $\geq 2$ $\ge 2$ with only odd elements: $2⁶ - 1 - 6 = 57$ $2^6 - 1 - 6 = 57$

Required count:

Total - (all odd subsets) - (even singletons) = 2¹¹ - 2⁶ - 5 = 2048 - 64 - 5 = 1979

Pattern Recognition

Complement method: Total subsets minus subsets containing only odd elements minus even singletons.

Chapter Mix

Class 11 Maths: Permutations and Combinations

The number of strictly increasing functions f$f$ from the set 1, 2, 3, 4, 5, 6$\{1, 2, 3, 4, 5, 6\}$ to the set 1, 2, 3, …, 9$\{1, 2, 3, \ldots, 9\}$ such that f(i) ≠ i$f(i) \neq i$ for 1 ≤ i ≤ 6$1 \leq i \leq 6$ , is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Case Analysis on f(1)

Step 2: Subsequent Cases

Step 3: Total Sum

Pattern Recognition

Chapter Mix

Reference Study Guides

More Permutations and Combinations Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Compute q

Step 2: Final Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Legendre's Formula

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Case Breakdown

Step 2: Calculating Combinations for Each Case

Step 3: Total Pentagons

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Complementary Subtraction

Pattern Recognition

Chapter Mix

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The number of strictly increasing functions $f$ $f$ from the set $1, 2, 3, 4, 5, 6$ $\{1, 2, 3, 4, 5, 6\}$ to the set $1, 2, 3, \dots, 9$ $\{1, 2, 3, \ldots, 9\}$ such that $f(i) \neq i$ $f(i) \neq i$ for $1 \leq i \leq 6$ $1 \leq i \leq 6$ , is equal to: