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

Related Formula

For a strictly increasing function f: A → B$f: A \to B$ where |A| = m$|A| = m$ and |B| = n$|B| = n$, the number of functions without restrictions is nm$\binom{n}{m}$.

Core Logic

We need strictly increasing functions f: 1,2,3,4,5,6 → 1,2, ,9$f: \{1,2,3,4,5,6\} \to \{1,2,\dots,9\}$ subject to f(i) ≠ i$f(i) \neq i$. Since f$f$ is strictly increasing, f(i) ≥ 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) ≠ i$f(i) \neq i$. Thus, f(i) > i$f(i) > i$ for all 1 ≤ i ≤ 6$1 \leq i \leq 6$. This implies f(1) ≥ 2$f(1) \geq 2$.

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) ≥ 3 > 2$f(2) \ge 3 > 2$). Number of ways = 75 = 21$= \binom{7}{5} = 21$.

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 = 65 = 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 = 55 = 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) ≠ 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.

Chapter Mix

Class 11 Maths: Permutations and Combinations Class 12 Maths: Functions

Related Formula

Modular arithmetic reductions for power cycles:

Core Logic

Analyze condition p$p$: (6^m + 9ⁿ)$(6^m + 9^n)$ is divisible by 5. 6 ≡ 1 5 ⇒ 6^m ≡ 1^m ≡ 1 5$6 \equiv 1 \pmod 5 \Rightarrow 6^m \equiv 1^m \equiv 1 \pmod 5$. 9 ≡ -1 5 ⇒ 9ⁿ ≡ (-1)ⁿ 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 ≡ 0 5 ⇒ (-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, , 50$m \in \{1, 2, \dots, 50\}$, m$m$ can be anything (50 choices). Since n$n$ must be odd in 1, , 50$\{1, \dots, 50\}$, n$n$ has 25 choices. p = 50 × 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 ≤ 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:

Explanation for counts: If m+n = S$m+n = S$, and m, n ≥ 1$m, n \geq 1$, the number of ways is S-1$S-1$ (since S ≤ 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

Pattern Recognition

Modular exponentiation immediately shrinks large powers to ± 1$\pm 1$. The sum m+n=S$m+n=S$ where 1 ≤ m,n ≤ N$1 \le m,n \le N$ has exactly S-1$S-1$ solutions if S ≤ N$S \le N$, allowing instant combinatorics tallying without manual counting.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Core Logic

To find the maximum power n$n$ such that 7ⁿ$7^n$ divides 101!$101!$, we need to find the exponent of the prime 7$7$ in the prime factorization of 101!$101!$.

Step 1: Apply Legendre's Formula

n = 14 + 2 + 0$n = 14 + 2 + 0$ n = 16$n = 16$

Pattern Recognition

For prime p$p$ in N!$N!$, iteratively divide N$N$ by p$p$ taking only integer parts and sum them. Fast mental math: 101 ÷ 7 = 14$101 \div 7 = 14$, 14 ÷ 7 = 2$14 \div 7 = 2$. 14+2=16$14+2=16$.

Chapter Mix

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

Related Formula

Core Logic

A pentagon requires 5 distinct points as vertices. No three points can be collinear to form a proper polygon. Since the available points lie on the three sides of a triangle, picking 3 points from the same side will form a degenerate straight line segment rather than building a strictly convex vertex frame.

Therefore, we must select the 5 points distributed across the three sides (AB with 4 points, BC with 5 points, AC with 4 points) such that a maximum of 2 points is selected from any one side.

Step 1: Case Breakdown

We need to choose 5 points total from the three groups (4, 5, 4) with the condition that no group contributes more than 2 points. The only valid numerical partitions of 5 into 3 parts bounded by 2 are:

• Case 1: 2 points from AB, 2 points from BC, 1 point from AC
• Case 2: 2 points from AB, 1 point from BC, 2 points from AC
• Case 3: 1 point from AB, 2 points from BC, 2 points from AC

Step 2: Calculating Combinations for Each Case

Case 1: (2 from AB, 2 from BC, 1 from AC)

Case 2: (2 from AB, 1 from BC, 2 from AC)

Case 3: (1 from AB, 2 from BC, 2 from AC)

Step 3: Total Pentagons

Total number of pentagons is the sum of all valid cases:

Pattern Recognition

For polygon formation from collinear sets, always frame it as a restricted partition problem. A polygon of k$k$ sides requires selecting k$k$ vertices such that no maximum allowable threshold of collinearity is breached (for a strict polygon, no 3 points on a line).

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Complementary counting principle:

Core Logic

Set S = 1, 2, 3, , 11$S = \{1, 2, 3, \dots, 11\}$ contains 11 elements (6 odd: 1, 3, 5, 7, 9, 11$\{1, 3, 5, 7, 9, 11\}$ and 5 even: 2, 4, 6, 8, 10$\{2, 4, 6, 8, 10\}$).

• Total possible subsets of S = 2¹¹ = 2048$S = 2^{11} = 2048$.
• Subsets where product is odd consist entirely of odd numbers: 2⁶ = 64$2^6 = 64$.
• Singletons with even product: 5$5$ (the even numbers themselves).

Step 1: Complementary Subtraction

Excluded subsets:

• Empty set (size 0): 1$1$
• Singletons with odd elements: 6$6$
• Singletons with even elements: 5$5$
• Subsets of size ≥ 2$\ge 2$ with only odd elements: 2⁶ - 1 - 6 = 57$2^6 - 1 - 6 = 57$

Required count:

Pattern Recognition

Complement method: Total subsets minus subsets containing only odd elements minus even singletons.

Chapter Mix

Class 11 Maths: Permutations and Combinations