The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that…

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 number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is

Solution & Explanation

Related Formula

Core Logic

Case 1: Single letter used 6 \times

Case 2: Two distinct letters used

Case 3: Three distinct letters used

Step 1: Calculate Total Words

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

More Permutations and Combinations Questions — jee_main_2025_29_jan_morning

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