### 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

### Related Formula Number of ways to choose 3$3$ points out of N$N$ points is ^N C_3$^N C_3$. If any points are collinear, choosing 3$3$ points from those collinear points will form a straight line instead of a triangle. ### Core Logic Total number of points = 1$1$ (origin O$O$) + 12$12$ (on L_1$L_1$) + 9$9$ (on L_2$L_2$) = 22$22$ points. Triangles are formed by choosing any 3$3$ points except those that are collinear. ### Step 1: Identifying Collinear Sets 1. The set of points lying on L_1$L_1$ includes O, P_1, dots, P_12$O, P_1, \dots, P_{12}$, which is 13$13$ collinear points. Number of collinear combinations = ^13 C_3$^{13} C_3$ 2. The set of points lying on L_2$L_2$ includes O, Q_1, dots, Q_9$O, Q_1, \dots, Q_9$, which is 10$10$ collinear points. Number of collinear combinations = ^10 C_3$^{10} C_3$ ### Step 2: Triangle Calculation Total Triangles = Total choices of 3 points - Collinear choices on L_1$L_1$ - Collinear choices on L_2$L_2$ textTotal = ^22 C_3 - ^13 C_3 - ^10 C_3$\text{Total } = ^{22} C_3 - ^{13} C_3 - ^{10} C_3$ Calculating individual combinations: - ^22 C_3 = frac22 times 21 times 203 times 2 times 1 = 1540$^{22} C_3 = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540$ - ^13 C_3 = frac13 times 12 times 113 times 2 times 1 = 286$^{13} C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$ - ^10 C_3 = frac10 times 9 times 83 times 2 times 1 = 120$^{10} C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ textTriangles = 1540 - 286 - 120 = 1134$\text{Triangles} = 1540 - 286 - 120 = 1134$ ### Pattern Recognition Alternatively, count using partition combinations to avoid large factorials: textTriangles = (^12 C_2 times ^9 C_1) + (^9 C_2 times ^12 C_1) + (1 times ^12 C_1 times ^9 C_1)$\text{Triangles} = (^{12} C_2 \times ^9 C_1) + (^9 C_2 \times ^{12} C_1) + (1 \times ^{12} C_1 \times ^9 C_1)$ = (66 times 9) + (36 times 12) + (108) = 594 + 432 + 108 = 1134$= (66 \times 9) + (36 \times 12) + (108) = 594 + 432 + 108 = 1134$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula Number of ways to select r$r$ items from a pool of n$n$ distinct objects: ^nC_r = fracn!r!(n-r)!$^nC_r = \frac{n!}{r!(n-r)!}$ ### Core Logic Total required players = 10$10$. Constraints: - Minimum 4 batsmen and minimum 4 bowlers. - 1 specific batsman (captain) and 1 specific bowler (vice-captain) are fixed (already selected). Remaining selection required: - Total players left to choose = 10 - 2 = 8$10 - 2 = 8$ players. - Available remaining pool: - Batsmen available = 7 - 1 = 6$7 - 1 = 6$ batsmen. - Bowlers available = 6 - 1 = 5$6 - 1 = 5$ bowlers. Adjusted structural constraints for the remaining 8 slots: - Needs at least 4 - 1 = 3$4 - 1 = 3$ more batsmen. - Needs at least 4 - 1 = 3$4 - 1 = 3$ more bowlers. ### Step 1: Set Up Case Combinations Let x$x$ be the number of additional batsmen and y$y$ be the number of additional bowlers selected, where x + y = 8$x + y = 8$ with x ge 3$x \ge 3$ and y ge 3$y \ge 3$. Possible case matches: - **Case 1**: 5$5$ batsmen and 3$3$ bowlers (x=5, y=3$x=5, y=3$) - **Case 2**: 4$4$ batsmen and 4$4$ bowlers (x=4, y=4$x=4, y=4$) - **Case 3**: 3$3$ batsmen and 5$5$ bowlers (x=3, y=5$x=3, y=5$) ### Step 2: Calculate Each Case Value 1. For Case 1: textWays = ^6C_5 times ^5C_3 = 6 times 10 = 60$\text{Ways} = ^6C_5 \times ^5C_3 = 6 \times 10 = 60$ 2. For Case 2: textWays = ^6C_4 times ^5C_4 = 15 times 5 = 75$\text{Ways} = ^6C_4 \times ^5C_4 = 15 \times 5 = 75$ 3. For Case 3: textWays = ^6C_3 times ^5C_5 = 20 times 1 = 20$\text{Ways} = ^6C_3 \times ^5C_5 = 20 \times 1 = 20$ ### Step 3: Compute Total Ways Sum the combinations across all valid exhaustive paths: textTotal Ways = 60 + 75 + 20 = 155$\text{Total Ways} = 60 + 75 + 20 = 155$ ### Pattern Recognition When specific roles (like captain/vice-captain) are strictly forced into the selection group, always remove them from both the operational choice pool size (n$n$) and the final destination requirement count (r$r$) before designing your target case distributions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations