Line L_1$L_1$ of slope 2$2$ and line L_2$L_2$ of slope frac12$\frac{1}{2}$ intersect at the origin O$O$. In the first quadrant, P_1, P_2, dots, P_12$P_1, P_2, \dots, P_{12}$ are 12$12$ points on line L_1$L_1$ and Q_1, Q_2, dots, Q_9$Q_1, Q_2, \dots, Q_9$ are 9$9$ points on line L_2$L_2$. Then the total number of triangles, that can be formed having vertices at three of the 22$22$ points O, P_1, P_2, dots, P_12, Q_1, Q_2, dots, Q_9$O, P_1, P_2, \dots, P_{12}, Q_1, Q_2, \dots, Q_9$, is:

### 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

### Related Formula The number of ways to choose r$r$ non-consecutive elements from a set of n$n$ items is given by the binomial formula: textWays = ^n - r + 1C_r$$\text{Ways} = {}^{n - r + 1}C_r$$ ### Core Logic We need to find the total number of subsets for n = 5$n = 5$ items without consecutive values. We break down the options based on the size of the subset (r$r$), ranging from an empty set (r=0$r=0$) up to the maximum possible non-consecutive size (r=3$r=3$). ### Step 1: Compute Combinations for Each Size 1. Subsets containing **no elements** (r = 0$r = 0$): textWays = 1 quad (textThe empty set emptyset)$$\text{Ways} = 1 \quad (\text{The empty set } \emptyset)$$ 2. Subsets containing **exactly 1 element** (r = 1$r = 1$): textWays = ^5 - 1 + 1C_1 = ^5C_1 = 5 quad (\1\, \2\, \3\, \4\, \5\)$$\text{Ways} = {}^{5 - 1 + 1}C_1 = {}^5C_1 = 5 \quad (\{1\}, \{2\}, \{3\}, \{4\}, \{5\})$$ 3. Subsets containing **exactly 2 elements** (r = 2$r = 2$): textWays = ^5 - 2 + 1C_2 = ^4C_2 = 6$$\text{Ways} = {}^{5 - 2 + 1}C_2 = {}^4C_2 = 6$$ 4. Subsets containing **exactly 3 elements** (r = 3$r = 3$): textWays = ^5 - 3 + 1C_3 = ^3C_3 = 1 quad (\1, 3, 5\)$$\text{Ways} = {}^{5 - 3 + 1}C_3 = {}^3C_3 = 1 \quad (\{1, 3, 5\})$$ ### Step 2: Sum the Total Counts Add all possible valid non-consecutive subset paths together: textTotal Subsets n(S_5) = 1 + 5 + 6 + 1 = 13$$\text{Total Subsets } n(S_5) = 1 + 5 + 6 + 1 = 13$$ ### Pattern Recognition Shortcut: The total number of non-consecutive subsets for a set of size n$n$ follows the Fibonacci sequence pattern: F_n+2$F_{n+2}$. For n=5$n=5$, the value is the 7th Fibonacci number, which is exactly 13$13$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula textNet Triangles = binomn3 - binomm3$$\text{Net Triangles} = \binom{n}{3} - \binom{m}{3}$$ ### Core Logic To calculate valid un-collapsed geometric triangles, find total ways to select 3 distinct coordinate indicators from the set and remove choices restricted entirely within the inline row sequence. ### Step 1: Compute Full Dynamic Combinations Choosing 3 general elements from the array size of 12: binom123 = frac12 times 11 times 103 times 2 times 1 = 220$$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$ ### Step 2: Isolate Internal Flattened Collinear Triplets Choosing 3 items completely bundled inside the collinear group row of size 5: binom53 = 10$$\binom{5}{3} = 10$$ ### Step 3: Deduce Triangles textTriangles Generated = 220 - 10 = 210$$\text{Triangles Generated} = 220 - 10 = 210$$ ### Pattern Recognition Points in a straight line cannot create spatial enclosing fields. Subtracting localized combinations from generalized permutations accounts for structural constraints. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula Number of permutations of n$n$ distinct letters is given by: textPermutations = n!$$\text{Permutations} = n!$$ ### Core Logic Sort the distinct letters of "KANPUR" alphabetically: textA, K, N, P, R, U$$\text{A, K, N, P, R, U}$$ Track alphabetical structural sets row by row: Words starting with A: 5! = 120$5! = 120$ Words starting with K: 5! = 120$5! = 120$ (Cumulative: 240) Words starting with N: 5! = 120$5! = 120$ (Cumulative: 360) ### Step 1: Parse the Next Character Layer We need the 440th word, so the next block begins with P: Words starting with PA: 4! = 24$4! = 24$ (Cumulative: 384) Words starting with PK: 4! = 24$4! = 24$ (Cumulative: 408) Words starting with PN: 4! = 24$4! = 24$ (Cumulative: 432) ### Step 2: Reach the Targeted Count Remaining difference to hit 440 is 440 - 432 = 8$440 - 432 = 8$ words. The next group starts with PR. Alphabetical listings within PR: Words starting with PRA: 3! = 6$3! = 6$ (Cumulative: 438) Now list alphabetically within PRK: 439th word: PRKANU 440th word: PRKAUN ### Pattern Recognition Factorial blocks eliminate large numbers quickly. Once the cumulative total lands within a narrow range of the answer, switch over to writing combinations down manually to ensure high accuracy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations