Let S$S$ be the set of the first 11 natural numbers. Then the number of elements in A = \B subseteq S : n(B) ge 2 text and the product of all elements of B text is even\$A = \{B \subseteq S : n(B) \ge 2 \text{ and the product of all elements of } B \text{ is even}\}$ is ____.

### 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 functions f: \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$f(1) \geq 2$.

### 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$= \binom{7}{5} = 21$.

### 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

### 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: 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

### Related Formula textLegendre's Formula: The exponent of a prime p text in N! text is given by: $$\text{Legendre's Formula: The exponent of a prime } p \text{ in } N! \text{ is given by: }$$ E_p(N!) = leftlfloor fracNp rightrfloor + leftlfloor fracNp^2 rightrfloor + leftlfloor fracNp^3 rightrfloor + ldots$$E_p(N!) = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \ldots$$ ### Core Logic To find the maximum power n$n$ such that 7^n$7^n$ 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 = leftlfloor frac1017 rightrfloor + leftlfloor frac1017^2 rightrfloor + leftlfloor frac1017^3 rightrfloor + ldots$$n = \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{7^2} \right\rfloor + \left\lfloor \frac{101}{7^3} \right\rfloor + \ldots$$ n = leftlfloor frac1017 rightrfloor + leftlfloor frac10149 rightrfloor + leftlfloor frac101343 rightrfloor$$n = \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{49} \right\rfloor + \left\lfloor \frac{101}{343} \right\rfloor$$ 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 div 7 = 14$101 \div 7 = 14$, 14 div 7 = 2$14 \div 7 = 2$. 14+2=16$14+2=16$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations Class 11 Maths: Number Theory

### Related Formula textCombinations ^nC_r = fracn!r!(n-r)!$$\text{Combinations } ^{n}C_{r} = \frac{n!}{r!(n-r)!}$$ ### 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) ^4C_2 times ^5C_2 times ^4C_1 = 6 times 10 times 4 = 240$$^{4}C_{2} \times ^{5}C_{2} \times ^{4}C_{1} = 6 \times 10 \times 4 = 240$$ **Case 2:** (2 from AB, 1 from BC, 2 from AC) ^4C_2 times ^5C_1 times ^4C_2 = 6 times 5 times 6 = 180$$^{4}C_{2} \times ^{5}C_{1} \times ^{4}C_{2} = 6 \times 5 \times 6 = 180$$ **Case 3:** (1 from AB, 2 from BC, 2 from AC) ^4C_1 times ^5C_2 times ^4C_2 = 4 times 10 times 6 = 240$$^{4}C_{1} \times ^{5}C_{2} \times ^{4}C_{2} = 4 \times 10 \times 6 = 240$$ ### Step 3: Total Pentagons Total number of pentagons is the sum of all valid cases: textTotal = 240 + 180 + 240 = 660$$\text{Total} = 240 + 180 + 240 = 660$$

### 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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations