The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is (1) 4608 (2) 5720 (3) 5719 (4) 4607

### 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