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) neq i for 1 leq i leq 6 , is equal to:

Solution & Explanation

### Related Formula For a strictly increasing function f: A to B where |A| = m and |B| = n, the number of functions without restrictions is binomnm. ### Core Logic We need strictly increasing functions f: \1,2,3,4,5,6\ to \1,2,dots,9\ subject to f(i) neq i. Since f is strictly increasing, f(i) geq i must always hold because the target values are drawn from an equally spaced domain. If f(i) = i for any i, it forces a strict ladder down to 1. But we are given f(i) neq i. Thus, f(i) > i for all 1 leq i leq 6. This implies f(1) geq 2.
Function mapping sets diagram Q9 - JEE Main 2026 Morning
Function mapping sets diagram Q9 - JEE Main 2026 Morning
### Step 1: Case Analysis on f(1) Since f(1) > 1, we evaluate possible starting points: Case 1: f(1) = 2 Remaining 5 values f(2) to f(6) must be strictly increasing and chosen from \3, 4, 5, 6, 7, 8, 9\ (7 available numbers). Since f(1)=2, f(i)>i is naturally preserved for subsequent elements (e.g., f(2) ge 3 > 2). Number of ways = binom75 = 21.
Function mapping sets diagram Q9 - JEE Main 2026 Morning
Function mapping sets diagram Q9 - JEE Main 2026 Morning
### Step 2: Subsequent Cases Case 2: f(1) = 3 Remaining 5 values chosen from \4, 5, 6, 7, 8, 9\ (6 available numbers). Number of ways = binom65 = 6. Case 3: f(1) = 4 Remaining 5 values chosen from \5, 6, 7, 8, 9\ (5 available numbers). Number of ways = binom55 = 1. Case 4: f(1) = 5 Requires choosing 5 values from \6,7,8,9\, which is impossible. ### Step 3: Total Sum Total number of valid functions = 21 + 6 + 1 = 28. ### Pattern Recognition For f(i) neq i on strictly increasing integer arrays, f(x) - x > 0. Using the substitution g(x) = f(x) - x, you convert a constrained increasing function into a standard non-decreasing one, or simply pivot on f(1) and sum the cascading binomials. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations Class 12 Maths: Functions

Reference Study Guides

Permutations And Combinations

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, dots, 50\ \ . If the number of elements (m, n) in S such that 6^m + 9^n is a multiple of 5 is p and the number of elements (m, n) in 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 pmodm Rightarrow a^k equiv b^k pmodm ### Core Logic Analyze condition p: (6^m + 9^n) is divisible by 5. 6 equiv 1 pmod 5 Rightarrow 6^m equiv 1^m equiv 1 pmod 5. 9 equiv -1 pmod 5 Rightarrow 9^n equiv (-1)^n pmod 5. For the sum to be divisible by 5: 1 + (-1)^n equiv 0 pmod 5 Rightarrow (-1)^n = -1. This implies n must be an ODD integer. Since m in \1, 2, dots, 50\, m can be anything (50 choices). Since n must be odd in \1, dots, 50\, n has 25 choices. p = 50 times 25 = 1250. ### Step 1: Compute q Analyze condition q: (m + n) is the square of a prime number. Max value of m+n = 50+50 = 100. Primes whose squares are leq 100: 2, 3, 5, 7. Their squares are 4, 9, 25, 49. So m+n can be 4, 9, 25, 49. Match List-I with List-II:
m+n=4m+n=9m+n=25m+n=49
No. of ways382448
Explanation for counts: If m+n = S, and m, n geq 1, the number of ways is S-1 (since S leq 50). For S=4: 3 ways. For S=9: 8 ways. For S=25: 24 ways. For S=49: 48 ways. 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. The sum m+n=S where 1 le m,n le N has exactly S-1 solutions if S le N, allowing instant combinatorics tallying without manual counting. ### Evaluation Rubric / Model Answer null ### 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 textNumber of arrangements of r text items in n text boxes = binomnr 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): textTotal unrestricted arrangements = binom85 cdot 5! = 56 cdot 120 = 6720 ### Step 2: Identify and subtract the invalid empty-row cases Let the rows be R_1, R_2, and 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 R_1 (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R_2 and R_3: textWays = binom55 cdot 5! = 120 - Case 2: Row R_3 (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R_1 and R_2: textWays = binom55 cdot 5! = 120 - Case 3: Row R_2 (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R_1 and R_3: textWays = binom65 cdot 5! = 6 cdot 120 = 720 ### Step 3: Calculate the final valid arrangements Subtracting all empty-row cases from the total arrangements: textValid 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. ### Evaluation Rubric / Model Answer null ### 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 mathbbN such that 3^n divides 50! is:
  • A. 21
  • B. 22
  • C. 20
  • D. 23

Solution

### Related Formula The exponent of a prime p in N! is given by Legendre's formula: E_p(N!) = left[fracNpright] + left[fracNp^2right] + left[fracNp^3right] + dots ### Core Logic To find the highest power of 3 that divides 50!, calculate the sum of the greatest integer functions for successive powers of 3 up to 50. ### Step 1: Computation Applying the formula for N = 50 and p = 3: E_3(50!) = left[frac503right] + left[frac509right] + left[frac5027right] + left[frac5081right] E_3(50!) = 16 + 5 + 1 + 0 = 22 ### Pattern Recognition Quickly divide by powers of 3: 50/3 rightarrow 16; 16/3 rightarrow 5; 5/3 rightarrow 1. Summing them up yields 16 + 5 + 1 = 22 instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations
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
  • B. 45
  • C. 2520
  • D. 1820

Solution

### Related Formula The number of permutations of n objects where p are of one kind, q are of another kind, and r are of a third kind is: textTotal Permutations = fracn!p! cdot q! cdot r! ### Core Logic The sequence has 10 terms chosen from \0, 1, 2\. It contains exactly five 1s and exactly three 2s. This leaves exactly 10 - 5 - 3 = 2 terms to be filled by 0s. ### Step 1: Arrangement Calculation We need to arrange five 1s, three 2s, and two 0s. The number of unique sequences is: textTotal Sequences = frac10!5! cdot 3! cdot 2! textTotal Sequences = frac10 times 9 times 8 times 7 times 63 times 2 times 1 times 2 times 1 = 2520 ### Pattern Recognition Note that sequences can start with 0 since it asks for general sequences of ten terms rather than a standard non-zero multi-digit number representation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

More Permutations and Combinations Questions — jee_main_2026_21_jan_morning

Practice all Permutations and Combinations previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...