The largest value of n, for which 40n divides 60!, is — jee_main_2026_24_january

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q jee_main_2025_02_april_morning Permutation of Multiset

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to:

A. $360$ $360$
B. $45$ $45$
C. $2520$ $2520$
D. $1820$ $1820$

Solution

Related Formula

The number of permutations of $n$ $n$ objects where $p$ $p$ are of one kind, $q$ $q$ are of another kind, and $r$ $r$ are of a third kind is:

Total Permutations = (n!)/(p! \cdot q! \cdot r!)

Core Logic

The sequence has $10$ $10$ terms chosen from $0, 1, 2$ $\{0, 1, 2\}$. It contains exactly five $1$ $1$s and exactly three $2$ $2$s. This leaves exactly $10 - 5 - 3 = 2$ $10 - 5 - 3 = 2$ terms to be filled by $0$ $0$s.

Step 1: Arrangement Calculation

We need to arrange five $1$ $1$s, three $2$ $2$s, and two $0$ $0$s. The number of unique sequences is:

Total Sequences = (10!)/(5! \cdot 3! \cdot 2!)

Total Sequences = (10 \times 9 \times 8 \times 7 \times 6)/(3 \times 2 \times 1 \times 2 \times 1) = 2520

Pattern Recognition

Note that sequences can start with $0$ $0$ since it asks for general sequences of ten terms rather than a standard non-zero multi-digit number representation.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q58 jee_main_2025_03_april_evening Geometry Problems

Line

L₁

of slope

2

and line

L₂

of slope

(1)/(2)

intersect at the origin

O

. In the first quadrant,

P₁, P₂, , P₁₂

are

12

points on line

L₁

and

Q₁, Q₂, , Q₉

are

9

points on line

L₂

. Then the total number of triangles, that can be formed having vertices at three of the

22

points

O, P₁, P₂, , P₁₂, Q₁, Q₂, , Q₉

, is:

A. $1080$ $1080$
B. $1134$ $1134$
C. $1026$ $1026$
D. $1188$ $1188$

Solution

Related Formula

Number of ways to choose $3$ $3$ points out of $N$ $N$ points is $^N C₃$ $^N C_3$. If any points are collinear, choosing $3$ $3$ points from those collinear points will form a straight line instead of a triangle.

Core Logic

Total number of points = $1$ $1$ (origin $O$ $O$) + $12$ $12$ (on $L₁$ $L_1$) + $9$ $9$ (on $L₂$ $L_2$) = $22$ $22$ points. Triangles are formed by choosing any $3$ $3$ points except those that are collinear.

Step 1: Identifying Collinear Sets

The set of points lying on $L₁$ $L_1$ includes $O, P₁, , P₁₂$ $O, P_1, \dots, P_{12}$, which is $13$ $13$ collinear points.

Number of collinear combinations = $¹³ C₃$ $^{13} C_3$

The set of points lying on $L₂$ $L_2$ includes $O, Q₁, , Q₉$ $O, Q_1, \dots, Q_9$, which is $10$ $10$ collinear points.

Number of collinear combinations = $¹⁰ C₃$ $^{10} C_3$

Step 2: Triangle Calculation

Total Triangles = Total choices of 3 points - Collinear choices on $L₁$ $L_1$ - Collinear choices on $L₂$ $L_2$

Total = ²² C₃ - ¹³ C₃ - ¹⁰ C₃

Calculating individual combinations:

$²² C₃ = (22 \times 21 \times 20)/(3 \times 2 \times 1) = 1540$
$¹³ C₃ = (13 \times 12 \times 11)/(3 \times 2 \times 1) = 286$
$¹⁰ C₃ = (10 \times 9 \times 8)/(3 \times 2 \times 1) = 120$

Triangles = 1540 - 286 - 120 = 1134

Pattern Recognition

Alternatively, count using partition combinations to avoid large factorials:

Triangles = (¹² C₂ \times ⁹ C₁) + (⁹ C₂ \times ¹² C₁) + (1 \times ¹² C₁ \times ⁹ C₁)

= (66 \times 9) + (36 \times 12) + (108) = 594 + 432 + 108 = 1134

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q57 jee_main_2025_07_april_morning Practical Problems on Combinations

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

A. $165$ $165$
B. $155$ $155$
C. $145$ $145$
D. $135$ $135$

Solution

Related Formula

Number of ways to select $r$ $r$ items from a pool of $n$ $n$ distinct objects:

ⁿCᵣ = (n!)/(r!(n-r)!)

Core Logic

Total required players = $10$ $10$. Constraints:

Minimum 4 batsmen and minimum 4 bowlers.
1 specific batsman (captain) and 1 specific bowler (vice-captain) are fixed (already selected).

Remaining selection required:

Total players left to choose = $10 - 2 = 8$ $10 - 2 = 8$ players.
Available remaining pool:
Batsmen available = $7 - 1 = 6$ $7 - 1 = 6$ batsmen.
Bowlers available = $6 - 1 = 5$ $6 - 1 = 5$ bowlers.

Adjusted structural constraints for the remaining 8 slots:

Needs at least $4 - 1 = 3$ $4 - 1 = 3$ more batsmen.
Needs at least $4 - 1 = 3$ $4 - 1 = 3$ more bowlers.

Step 1: Set Up Case Combinations

Let $x$ $x$ be the number of additional batsmen and $y$ $y$ be the number of additional bowlers selected, where $x + y = 8$ $x + y = 8$ with $x \geq 3$ $x \ge 3$ and $y \geq 3$ $y \ge 3$.

Possible case matches:

Case 1: $5$ $5$ batsmen and $3$ $3$ bowlers ( $x=5, y=3$ $x=5, y=3$)
Case 2: $4$ $4$ batsmen and $4$ $4$ bowlers ( $x=4, y=4$ $x=4, y=4$)
Case 3: $3$ $3$ batsmen and $5$ $5$ bowlers ( $x=3, y=5$ $x=3, y=5$)

Step 2: Calculate Each Case Value

For Case 1:

Ways = ⁶C₅ \times ⁵C₃ = 6 \times 10 = 60

For Case 2:

Ways = ⁶C₄ \times ⁵C₄ = 15 \times 5 = 75

For Case 3:

Ways = ⁶C₃ \times ⁵C₅ = 20 \times 1 = 20

Step 3: Compute Total Ways

Sum the combinations across all valid exhaustive paths:

Total Ways = 60 + 75 + 20 = 155

Pattern Recognition

When specific roles (like captain/vice-captain) are strictly forced into the selection group, always remove them from both the operational choice pool size ( $n$ $n$) and the final destination requirement count ( $r$ $r$) before designing your target case distributions.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q75 jee_main_2025_07_april_morning Subsets without Consecutive Elements

For

n \geq 2

, let

Sₙ

denote the set of all subsets of

1, 2, , n

with no two consecutive numbers. For example

1, 3, 5 in S₆

, but

1, 2, 4 S₆

. Then

n(S₅)

is equal to

Numerical Answer. Answer: 13 to 13

Solution

Related Formula

The number of ways to choose $r$ $r$ non-consecutive elements from a set of $n$ $n$ items is given by the binomial formula:

Ways = n - r + 1Cᵣ

Core Logic

We need to find the total number of subsets for $n = 5$ $n = 5$ items without consecutive values. We break down the options based on the size of the subset ( $r$ $r$), ranging from an empty set ( $r=0$ $r=0$) up to the maximum possible non-consecutive size ( $r=3$ $r=3$).

Step 1: Compute Combinations for Each Size

Subsets containing no elements ( $r = 0$ $r = 0$):

Ways = 1 (The empty set )

Subsets containing exactly 1 element ( $r = 1$ $r = 1$):

Ways = 5 - 1 + 1C₁ = ⁵C₁ = 5 (1, 2, 3, 4, 5)

Subsets containing exactly 2 elements ( $r = 2$ $r = 2$):

Ways = 5 - 2 + 1C₂ = ⁴C₂ = 6

Subsets containing exactly 3 elements ( $r = 3$ $r = 3$):

Ways = 5 - 3 + 1C₃ = ³C₃ = 1 (1, 3, 5)

Step 2: Sum the Total Counts

Add all possible valid non-consecutive subset paths together:

Total Subsets n(S₅) = 1 + 5 + 6 + 1 = 13

Pattern Recognition

Shortcut: The total number of non-consecutive subsets for a set of size $n$ $n$ follows the Fibonacci sequence pattern: $Fₙ₊₂$ $F_{n+2}$. For $n=5$ $n=5$, the value is the 7th Fibonacci number, which is exactly $13$ $13$.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q63 jee_main_2025_08_april_evening Geometry Problems in Combinatorics

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

A. $230$ $230$
B. $220$ $220$
C. $200$ $200$
D. $210$ $210$

Solution

Related Formula

Net Triangles = n3 - m3

Core Logic

To calculate valid un-collapsed geometric triangles, find total ways to select 3 distinct coordinate indicators from the set and remove choices restricted entirely within the inline row sequence.

Step 1: Compute Full Dynamic Combinations

Choosing 3 general elements from the array size of 12:

123 = (12 \times 11 \times 10)/(3 \times 2 \times 1) = 220

Step 2: Isolate Internal Flattened Collinear Triplets

Choosing 3 items completely bundled inside the collinear group row of size 5:

53 = 10

Step 3: Deduce Triangles

Triangles Generated = 220 - 10 = 210

Pattern Recognition

Points in a straight line cannot create spatial enclosing fields. Subtracting localized combinations from generalized permutations accounts for structural constraints.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

The largest value of n, for which 40ⁿ$40^{n}$ divides 60!$60!$, is

Solution & Explanation

Related Formula

Core Logic

Step 1: Exponent of 2 in 60!

Step 2: Exponent of 5 in 60!

Step 3: Calculating Limiting Factor

Pattern Recognition

Chapter Mix

Reference Study Guides

More Permutations and Combinations Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Arrangement Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identifying Collinear Sets

Step 2: Triangle Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up Case Combinations

Step 2: Calculate Each Case Value

Step 3: Compute Total Ways

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Combinations for Each Size

Step 2: Sum the Total Counts

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Full Dynamic Combinations

Step 2: Isolate Internal Flattened Collinear Triplets

Step 3: Deduce Triangles

Pattern Recognition

Chapter Mix

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The largest value of n, for which $40ⁿ$ $40^{n}$ divides $60!$ $60!$, is