Let S = p1, p2, ldots, p10 be the set of the first ten prime numbers. Let A = S P, where P is the set of all possible…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q10 jee_main_2026_24_january_evening Exponent of Prime in n!

The largest value of n, for which

40ⁿ

divides

60!

, is

A. $13$ $13$
B. $11$ $11$
C. $12$ $12$
D. $14$ $14$

Solution

Related Formula

Legendre's Formula for exponent of prime p in n!:

Eₚ(n!) = [ (n)/(p) ] + [ (n)/(p²) ] + [ (n)/(p³) ] +

where $[\cdot]$ $[\cdot]$ represents the greatest integer function.

Core Logic

To find the highest power of $40$ $40$ dividing $60!$ $60!$, we first prime factorize $40$ $40$.

40 = 2³ \times 5

Thus, $40ⁿ = 2³ⁿ \times 5ⁿ$ $40^n = 2^{3n} \times 5^n$. We need to find the exponents of $2$ $2$ and $5$ $5$ in $60!$ $60!$ and restrict $n$ $n$ to satisfy both components simultaneously.

Step 1: Exponent of 2 in 60!

E₂(60!) = [(60)/(2)] + [(60)/(4)] + [(60)/(8)] + [(60)/(16)] + [(60)/(32)]

= 30 + 15 + 7 + 3 + 1 = 56

Thus, $60!$ $60!$ contains $2⁵⁶$ $2^{56}$.

Step 2: Exponent of 5 in 60!

E₅(60!) = [(60)/(5)] + [(60)/(25)]

$= 12 + 2 = 14$ $= 12 + 2 = 14$

Thus, $60!$ $60!$ contains $5¹⁴$ $5^{14}$.

Step 3: Calculating Limiting Factor

From the factors, $60!$ $60!$ can be written as $2⁵⁶ \times 5¹⁴ \times K$ $2^{56} \times 5^{14} \times K$.

We need to construct factors of $40 = 2³ \times 5$ $40 = 2^3 \times 5$. From $2⁵⁶$ $2^{56}$, we can form $(2³)¹⁸$ $(2^3)^{18}$ with a remainder, so $2³$ $2^3$ limits at $18$ $18$. From $5¹⁴$ $5^{14}$, we can form $5¹⁴$ $5^{14}$, so $5$ $5$ limits at $14$ $14$.

The limiting factor is the exponent of $5$ $5$, which is $14$ $14$.

Therefore, the maximum value of $n$ $n$ is $14$ $14$.

Pattern Recognition

For a composite base $C = p₁a₁ p₂a₂$ $C = p_1^{a_1} p_2^{a_2}$, the highest power is $( [ Ep₁(n!)a₁ ], [ Ep₂(n!)a₂ ] )$ . Generally, the larger prime (here $5$ $5$) dictates the bottleneck.

Chapter Mix

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

Q7 jee_main_2026_28_january_morning Arrangement of Digits

Let

S = 1, 2, 3, 4, 5, 6, 7, 8, 9

. Let

x

be the number of 9-digit numbers formed using the digits of the set

S

such that only one digit is repeated and it is repeated exactly twice. Let

y

be the number of 9-digit numbers formed using the digits of the set

S

such that only two digits are repeated and each of these is repeated exactly twice. Then,

A. $29x = 5y$ $29x = 5y$
B. $45x = 7y$ $45x = 7y$
C. $21x = 4y$ $21x = 4y$
D. $56x = 9y$ $56x = 9y$

Solution

Core Logic

$S = 1, 2, 3, , 9$ $S = \{1, 2, 3, \dots, 9\}$ For $x$ $x$: 9-digit number where ONLY ONE digit is repeated exactly twice. This means the number contains 8 distinct digits from $S$ $S$. One digit appears twice, and 7 digits appear once. Total length = $2 + 7 = 9$ $2 + 7 = 9$. Calculation for $x$ $x$:

Select 1 digit to be repeated twice: $⁹C₁$ $^{9}C_1$
Select 7 digits out of the remaining 8: $⁸C₇$ $^{8}C_7$
Arrange these 9 digits (where 2 are identical): $(9!)/(2!)$ $\frac{9!}{2!}$

x = ⁹C₁ \cdot ⁸C₇ \cdot (9!)/(2) = (9 \times 8 \times 9!)/(2)

Step 1: Determine y

For $y$ $y$: 9-digit number where EXACTLY TWO digits are repeated twice each. This means we select 2 digits to appear twice, and we need $9 - 4 = 5$ $9 - 4 = 5$ more digits (which must all be distinct and appear once). Total length = $2 + 2 + 5 = 9$ $2 + 2 + 5 = 9$. Calculation for $y$ $y$:

Select 2 digits to be repeated twice: $⁹C₂$ $^{9}C_2$
Select 5 digits out of the remaining 7: $⁷C₅$ $^{7}C_5$
Arrange these 9 digits (two pairs of identicals): $(9!)/(2! \times 2!)$ $\frac{9!}{2! \times 2!}$

y = ⁹C₂ \cdot ⁷C₅ \cdot (9!)/(2! \times 2!) = (9 \times 8)/(2) \times (7 \times 6)/(2) \times (9!)/(4)

Step 2: Ratio of x and y

(x)/(y) = ((9 \times 8 \times 9!)/(2))/((9 \times 8 \times 7 \times 6 \times 9!)/(2 \times 2 \times 4)) = ((9 \times 8)/(2))/((9 \times 8)/(2) \times (7 \times 6)/(2) \times (1)/(2))

Simplifying the fraction:

(x)/(y) = (1)/((42)/(4)) = (4)/(42) = (2)/(21) (wait, let me re-evaluate)

Let's carefully compute: $x = (72)/(2) \times 9! = 36 \times 9!$ $y = 36 \times 21 \times (9!)/(4) = 9 \times 21 \times 9! = 189 \times 9!$

(x)/(y) = (36)/(189) = (4)/(21)

Step 3: Final Relation

(x)/(y) = (4)/(21) 21x = 4y

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q22 jee_main_2026_28_january_evening Distribution of Distinct Objects

Three persons enter in a lift at the ground floor. The lift will go up to

10th

floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to

Numerical Answer. Answer: 210 to 210

Solution

Core Logic

The lift can go to floors 1 through 10. It does not stop at 1, 2, or 3. The available floors for exit are 4, 5, 6, 7, 8, 9, 10. Number of available floors $n = 7$ $n = 7$. The three persons must exit at three different floors. So we must choose 3 distinct floors from the 7 available, and assign them to the 3 distinct persons.

Execution

Ways to choose 3 floors from 7: $⁷C₃$ $^{7}C_{3}$. Ways to arrange 3 persons on these 3 floors: $3!$ $3!$.

Total ways $= ⁷C₃ \times 3!$ $= ^{7}C_{3} \times 3!$

= (7 \times 6 \times 5)/(3 \times 2 \times 1) \times 6 = 35 \times 6 = 210

Pattern Recognition

Selecting sets for distinct people inherently combines combination selection with factorial permutation ( $ⁿCᵣ \times r!$ $^nC_r \times r!$ or directly $ⁿPᵣ$ $^nP_r$). Restricting available floors simply reduces $n$ $n$.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q66 jee_main_2025_02_april_evening Arrangements

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

Boxes layout diagram for Q66 - JEE Main 2025 Evening — The grid diagram shows 8 boxes arranged in three horizontal rows of sizes 3, 2, and 3.

A. 5880
B. 960
C. 840
D. 5760

Solution

Related Formula

Number of arrangements of r items in n boxes = nr \cdot r!

Core Logic

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty.

Step 1: Compute total unrestricted arrangements

The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E):

Total unrestricted arrangements = 85 \cdot 5! = 56 \cdot 120 = 6720

Step 2: Identify and subtract the invalid empty-row cases

Let the rows be $R₁$ $R_1$, $R₂$ $R_2$, and $R₃$ $R_3$, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at most 3 boxes, which cannot fit 5 letters). Thus, we only subtract cases where exactly one row is empty:

Case 1: Row $R₁$ $R_1$ (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of $R₂$ $R_2$ and $R₃$ $R_3$:

Ways = 55 \cdot 5! = 120

Case 2: Row $R₃$ $R_3$ (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of $R₁$ $R_1$ and $R₂$ $R_2$:

Ways = 55 \cdot 5! = 120

Case 3: Row $R₂$ $R_2$ (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of $R₁$ $R_1$ and $R₃$ $R_3$:

Ways = 65 \cdot 5! = 6 \cdot 120 = 720

Step 3: Calculate the final valid arrangements

Subtracting all empty-row cases from the total arrangements:

Valid arrangements = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760

Pattern Recognition

Inclusion-Exclusion Principle: For distribution problems with simple boundary exclusions, subtracting the complement set (invalid configurations) is mathematically much cleaner than calculating all possible partitions of row assignments.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q jee_main_2025_02_april_morning Exponent of Prime in a Factorial

The largest

n in N

such that

3ⁿ

divides

50!

is:

A. $21$ $21$
B. $22$ $22$
C. $20$ $20$
D. $23$ $23$

Solution

Related Formula

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

Eₚ(N!) = [(N)/(p)] + [(N)/(p²)] + [(N)/(p³)] +

Core Logic

To find the highest power of $3$ $3$ that divides $50!$ $50!$, calculate the sum of the greatest integer functions for successive powers of $3$ $3$ up to $50$ $50$.

Step 1: Computation

Applying the formula for $N = 50$ $N = 50$ and $p = 3$ $p = 3$:

E₃(50!) = [(50)/(3)] + [(50)/(9)] + [(50)/(27)] + [(50)/(81)]

E₃(50!) = 16 + 5 + 1 + 0 = 22

Pattern Recognition

Quickly divide by powers of $3$ $3$: $50/3 arrow 16$ $50/3 \rightarrow 16$; $16/3 arrow 5$ $16/3 \rightarrow 5$; $5/3 arrow 1$ $5/3 \rightarrow 1$. Summing them up yields $16 + 5 + 1 = 22$ $16 + 5 + 1 = 22$ instantly.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Solution & Explanation

Related Formula

Core Logic

Step 1: Count elements belonging to product set P

Step 2: Combine and Evaluate Total Ordered Pairs

Pattern Recognition

Chapter Mix

More Permutations and Combinations Previous-Year Questions — Page 3

Solution

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

Solution

Core Logic

Step 1: Determine y

Step 2: Ratio of x and y

Step 3: Final Relation

Chapter Mix

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute total unrestricted arrangements

Step 2: Identify and subtract the invalid empty-row cases

Step 3: Calculate the final valid arrangements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Computation

Pattern Recognition

Chapter Mix

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