In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is
Numerical Answer Type:
Enter a numerical valueAnswer: 11376 to 11376+4 marks
Solution & Explanation
### Related Formula
textCombinations: ^nC_r = fracn!r!(n-r)!$$\text{Combinations: } {}^nC_r = \frac{n!}{r!(n-r)!}$$
### Core Logic
Total Questions = 20 (A: 8, B: 6, C: 6).
Total to attempt = 15.
Minimum required from each section = 4.
Base attempt gives: 4 (textfrom A) + 4 (textfrom B) + 4 (textfrom C) = 12$4 (\text{from A}) + 4 (\text{from B}) + 4 (\text{from C}) = 12$ questions.
We have to distribute the remaining 15 - 12 = 3$15 - 12 = 3$ questions across the sections A, B, and C.
Let the additional questions picked be x, y, z$x, y, z$ for sections A, B, C respectively.
Then x + y + z = 3$x + y + z = 3$, with constraints based on the maximum questions per section:
A max extra = 8 - 4 = 4 Rightarrow x le 4$8 - 4 = 4 \Rightarrow x \le 4$
B max extra = 6 - 4 = 2 Rightarrow y le 2$6 - 4 = 2 \Rightarrow y \le 2$
C max extra = 6 - 4 = 2 Rightarrow z le 2$6 - 4 = 2 \Rightarrow z \le 2$
### Step 1: Identifying Valid Selection Cases
The possible sets of (x, y, z)$(x, y, z)$ are:
Case 1: (1, 1, 1)$(1, 1, 1)$Rightarrow$\Rightarrow$ Total picks: A(5), B(5), C(5)
Case 2: (2, 1, 0)$(2, 1, 0)$ and its permutations (respecting constraints).
Valid permutations:
- A gets 2, B gets 1, C gets 0 Rightarrow$\Rightarrow$ A(6), B(5), C(4)
- A gets 2, C gets 1, B gets 0 Rightarrow$\Rightarrow$ A(6), B(4), C(5)
- B gets 2, A gets 1, C gets 0 Rightarrow$\Rightarrow$ A(5), B(6), C(4)
- C gets 2, A gets 1, B gets 0 Rightarrow$\Rightarrow$ A(5), B(4), C(6)
(Note: B(2), C(1) or C(2), B(1) are not allowed if it forces A to take 0, wait, A gets 0 means A(4). A is allowed to have 4.) Let's check permutations of (2, 1, 0)$(2, 1, 0)$:
- A(4), B(6), C(5) [x=0, y=2, z=1]
- A(4), B(5), C(6) [x=0, y=1, z=2]
Case 3: (3, 0, 0)$(3, 0, 0)$ and permutations.
Since y le 2$y \le 2$ and z le 2$z \le 2$, only x$x$ can be 3.
So, x=3, y=0, z=0$x=3, y=0, z=0$Rightarrow$\Rightarrow$ A(7), B(4), C(4).
### Step 2: Calculating Combinations per Case
Let's list all valid final section breakdowns (A, B, C)$(A, B, C)$:
1) (5, 5, 5) Rightarrow ^8C_5 cdot ^6C_5 cdot ^6C_5 = 56 cdot 6 cdot 6 = 2016$(5, 5, 5) \Rightarrow {}^8C_5 \cdot {}^6C_5 \cdot {}^6C_5 = 56 \cdot 6 \cdot 6 = 2016$
2) (6, 5, 4) Rightarrow ^8C_6 cdot ^6C_5 cdot ^6C_4 = 28 cdot 6 cdot 15 = 2520$(6, 5, 4) \Rightarrow {}^8C_6 \cdot {}^6C_5 \cdot {}^6C_4 = 28 \cdot 6 \cdot 15 = 2520$
3) (6, 4, 5) Rightarrow ^8C_6 cdot ^6C_4 cdot ^6C_5 = 28 cdot 15 cdot 6 = 2520$(6, 4, 5) \Rightarrow {}^8C_6 \cdot {}^6C_4 \cdot {}^6C_5 = 28 \cdot 15 \cdot 6 = 2520$
4) (5, 6, 4) Rightarrow ^8C_5 cdot ^6C_6 cdot ^6C_4 = 56 cdot 1 cdot 15 = 840$(5, 6, 4) \Rightarrow {}^8C_5 \cdot {}^6C_6 \cdot {}^6C_4 = 56 \cdot 1 \cdot 15 = 840$
5) (5, 4, 6) Rightarrow ^8C_5 cdot ^6C_4 cdot ^6C_6 = 56 cdot 15 cdot 1 = 840$(5, 4, 6) \Rightarrow {}^8C_5 \cdot {}^6C_4 \cdot {}^6C_6 = 56 \cdot 15 \cdot 1 = 840$
6) (4, 6, 5) Rightarrow ^8C_4 cdot ^6C_6 cdot ^6C_5 = 70 cdot 1 cdot 6 = 420$(4, 6, 5) \Rightarrow {}^8C_4 \cdot {}^6C_6 \cdot {}^6C_5 = 70 \cdot 1 \cdot 6 = 420$
7) (4, 5, 6) Rightarrow ^8C_4 cdot ^6C_5 cdot ^6C_6 = 70 cdot 6 cdot 1 = 420$(4, 5, 6) \Rightarrow {}^8C_4 \cdot {}^6C_5 \cdot {}^6C_6 = 70 \cdot 6 \cdot 1 = 420$
8) (7, 4, 4) Rightarrow ^8C_7 cdot ^6C_4 cdot ^6C_4 = 8 cdot 15 cdot 15 = 1800$(7, 4, 4) \Rightarrow {}^8C_7 \cdot {}^6C_4 \cdot {}^6C_4 = 8 \cdot 15 \cdot 15 = 1800$
### Step 3: Summing the Total Ways
Total ways = 2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$
Total ways = 2016 + 5040 + 1680 + 840 + 1800 = 11376$2016 + 5040 + 1680 + 840 + 1800 = 11376$
### Pattern Recognition
Combinatorial distribution with rigid lower bounds is solved by shifting the baseline. Allocate the minimums immediately (4+4+4=12$4+4+4=12$), then distribute the remaining items via casework ensuring upper capacities aren't breached.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Permutations and Combinations
Keywords:#Combinations constraints#JEE Main 2024 Evening Q26#Permutations and Combinations JEE Main 2024#Selection of Objects JEE Main 2024
More Permutations and Combinations Previous-Year Questions
Q9jee_main_2026_21_jan_morningStrictly Increasing Functions and Derangements
The number of strictly increasing functionsf$f$ from the set \1, 2, 3, 4, 5, 6\$\{1, 2, 3, 4, 5, 6\}$ to the set \1, 2, 3, ldots, 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:
A. 21
B. 27
C. 22
D. 28
Solution
### Related Formula
For a strictly increasing function f: A to B$f: A \to B$ where |A| = m$|A| = m$ and |B| = n$|B| = n$, the number of functions without restrictions is binomnm$\binom{n}{m}$.
### Core Logic
We need strictly increasing functionsf: \1,2,3,4,5,6\ to \1,2,dots,9\$f: \{1,2,3,4,5,6\} \to \{1,2,\dots,9\}$ subject to f(i) neq i$f(i) \neq i$.
Since f$f$ is strictly increasing, f(i) geq i$f(i) \geq i$ must always hold because the target values are drawn from an equally spaced domain. If f(i) = i$f(i) = i$ for any i$i$, it forces a strict ladder down to 1. But we are given f(i) neq i$f(i) \neq i$. Thus, f(i) > i$f(i) > i$ for all 1 leq i leq 6$1 \leq i \leq 6$.
This implies f(1) geq 2$f(1) \geq 2$.
Function mapping sets diagram Q9 - JEE Main 2026 Morning
### Step 1: Case Analysis on f(1)
Since f(1) > 1$f(1) > 1$, we evaluate possible starting points:
Case 1: f(1) = 2$f(1) = 2$
Remaining 5 values f(2)$f(2)$ to f(6)$f(6)$ must be strictly increasing and chosen from \3, 4, 5, 6, 7, 8, 9\$\{3, 4, 5, 6, 7, 8, 9\}$ (7 available numbers). Since f(1)=2$f(1)=2$, f(i)>i$f(i)>i$ is naturally preserved for subsequent elements (e.g., f(2) ge 3 > 2$f(2) \ge 3 > 2$).
Number of ways = binom75 = 21$= \binom{7}{5} = 21$.
Function mapping sets diagram Q9 - JEE Main 2026 Morning
### Step 2: Subsequent Cases
Case 2: f(1) = 3$f(1) = 3$
Remaining 5 values chosen from \4, 5, 6, 7, 8, 9\$\{4, 5, 6, 7, 8, 9\}$ (6 available numbers).
Number of ways = binom65 = 6$= \binom{6}{5} = 6$.
Case 3: f(1) = 4$f(1) = 4$
Remaining 5 values chosen from \5, 6, 7, 8, 9\$\{5, 6, 7, 8, 9\}$ (5 available numbers).
Number of ways = binom55 = 1$= \binom{5}{5} = 1$.
Case 4: f(1) = 5$f(1) = 5$
Requires choosing 5 values from \6,7,8,9\$\{6,7,8,9\}$, which is impossible.
### Step 3: Total Sum
Total number of valid functions = 21 + 6 + 1 = 28$21 + 6 + 1 = 28$.
### Pattern Recognition
For f(i) neq i$f(i) \neq i$ on strictly increasing integer arrays, f(x) - x > 0$f(x) - x > 0$. Using the substitution g(x) = f(x) - x$g(x) = f(x) - x$, you convert a constrained increasing function into a standard non-decreasing one, or simply pivot on f(1)$f(1)$ and sum the cascading binomials.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Permutations and Combinations
Class 12 Maths: Functions
Q23jee_main_2026_21_jan_morningDivisibility and Counting Rules
Let S = \(m, n): m, n in \1, 2, 3, dots, 50\ \$S = \{(m, n): m, n \in \{1, 2, 3, \dots, 50\} \}$ . If the number of elements (m, n)$(m, n)$ in S$S$ such that 6^m + 9^n$6^m + 9^n$ is a multiple of 5 is p$p$ and the number of elements (m, n)$(m, n)$ in S$S$ such that m + n$m + n$ is a square of a prime number is q$q$ , then p + q$p + q$ 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$$a \equiv b \pmod{m} \Rightarrow a^k \equiv b^k \pmod{m}$$
### Core Logic
Analyze condition p$p$: (6^m + 9^n)$(6^m + 9^n)$ is divisible by 5.
6 equiv 1 pmod 5 Rightarrow 6^m equiv 1^m equiv 1 pmod 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$9 \equiv -1 \pmod 5 \Rightarrow 9^n \equiv (-1)^n \pmod 5$.
For the sum to be divisible by 5:
1 + (-1)^n equiv 0 pmod 5 Rightarrow (-1)^n = -1$1 + (-1)^n \equiv 0 \pmod 5 \Rightarrow (-1)^n = -1$.
This implies n$n$ must be an ODD integer.
Since m in \1, 2, dots, 50\$m \in \{1, 2, \dots, 50\}$, m$m$ can be anything (50 choices).
Since n$n$ must be odd in \1, dots, 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$$p + q = 1250 + 83 = 1333$$
### Pattern Recognition
Modular exponentiation immediately shrinks large powers to pm 1$\pm 1$. The sum m+n=S$m+n=S$ where 1 le m,n le N$1 \le m,n \le N$ has exactly S-1$S-1$ solutions if S le N$S \le N$, allowing instant combinatorics tallying without manual counting.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Permutations and Combinations
Q66jee_main_2025_02_april_eveningArrangements
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:
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!$$\text{Number of arrangements of } r \text{ items in } n \text{ boxes} = \binom{n}{r} \cdot r!$$
### 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$$\text{Total unrestricted arrangements} = \binom{8}{5} \cdot 5! = 56 \cdot 120 = 6720$$
### Step 2: Identify and subtract the invalid empty-row cases
Let the rows be R_1$R_1$, R_2$R_2$, and R_3$R_3$, 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$R_1$ (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R_2$R_2$ and R_3$R_3$:
textWays = binom55 cdot 5! = 120$$\text{Ways} = \binom{5}{5} \cdot 5! = 120$$
- Case 2: Row R_3$R_3$ (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R_1$R_1$ and R_2$R_2$:
textWays = binom55 cdot 5! = 120$$\text{Ways} = \binom{5}{5} \cdot 5! = 120$$
- Case 3: Row R_2$R_2$ (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R_1$R_1$ and R_3$R_3$:
textWays = binom65 cdot 5! = 6 cdot 120 = 720$$\text{Ways} = \binom{6}{5} \cdot 5! = 6 \cdot 120 = 720$$
### 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$$\text{Valid arrangements} = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760$$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Qjee_main_2025_02_april_morningExponent of Prime in a Factorial
The largest n in mathbbN$n \in \mathbb{N}$ such that 3^n$3^n$ divides 50!$50!$ 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:
E_p(N!) = left[fracNpright] + left[fracNp^2right] + left[fracNp^3right] + dots$$E_p(N!) = \left[\frac{N}{p}\right] + \left[\frac{N}{p^2}\right] + \left[\frac{N}{p^3}\right] + \dots$$
### 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$p = 3$:
E_3(50!) = left[frac503right] + left[frac509right] + left[frac5027right] + left[frac5081right]$$E_3(50!) = \left[\frac{50}{3}\right] + \left[\frac{50}{9}\right] + \left[\frac{50}{27}\right] + \left[\frac{50}{81}\right]$$E_3(50!) = 16 + 5 + 1 + 0 = 22$$E_3(50!) = 16 + 5 + 1 + 0 = 22$$
### Pattern Recognition
Quickly divide by powers of 3$3$: 50/3 rightarrow 16$50/3 \rightarrow 16$; 16/3 rightarrow 5$16/3 \rightarrow 5$; 5/3 rightarrow 1$5/3 \rightarrow 1$. Summing them up yields 16 + 5 + 1 = 22$16 + 5 + 1 = 22$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Qjee_main_2025_02_april_morningPermutation 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:
textTotal Permutations = fracn!p! cdot q! cdot r!$$\text{Total Permutations} = \frac{n!}{p! \cdot q! \cdot r!}$$
### 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:
textTotal Sequences = frac10!5! cdot 3! cdot 2!$$\text{Total Sequences} = \frac{10!}{5! \cdot 3! \cdot 2!}$$textTotal Sequences = frac10 times 9 times 8 times 7 times 63 times 2 times 1 times 2 times 1 = 2520$$\text{Total Sequences} = \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1} = 2520$$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
More Permutations and Combinations Questions — jee_main_2024_30_january_evening
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