From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, 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 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

### 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

### 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