### 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

### Related Formula For a 5-digit number, total permutations with repetition allowed for n$n$ digits is given by: textTotal Cases = d_1 times d_2 times d_3 times d_4 times d_5$\text{Total Cases} = d_1 \times d_2 \times d_3 \times d_4 \times d_5$ ### Core Logic We need 5-digit numbers greater than 50000 using digits \0, 1, 2, 3, 4, 5, 6, 7\$\{0, 1, 2, 3, 4, 5, 6, 7\}$ under the restriction d_1 + d_5 le 8$d_1 + d_5 \le 8$. Let's analyze the pairs (d_1, d_5)$(d_1, d_5)$ where d_1 in \5, 6, 7\$d_1 \in \{5, 6, 7\}$: Case I: d_1 = 5 Rightarrow d_5 in \0, 1, 2, 3\$d_1 = 5 \Rightarrow d_5 \in \{0, 1, 2, 3\}$ (4 options) Case II: d_1 = 6 Rightarrow d_5 in \0, 1, 2\$d_1 = 6 \Rightarrow d_5 \in \{0, 1, 2\}$ (3 options) Case III: d_1 = 7 Rightarrow d_5 in \0, 1\$d_1 = 7 \Rightarrow d_5 \in \{0, 1\}$ (2 options) Total choices for the first and last digits combined = 4 + 3 + 2 = 9$4 + 3 + 2 = 9$ pairs. ### Step 1: Calculating Intermediate Choices The middle three digits (d_2, d_3, d_4$d_2, d_3, d_4$) have no restrictions and can each be chosen from any of the 8 available digits. textNumber of ways = 9 times (8 times 8 times 8) = 4608$\text{Number of ways} = 9 \times (8 \times 8 \times 8) = 4608$ ### Step 2: Subtracting Boundary Conditions Since the question specifies numbers strictly greater than 50000, we must check if 50000 is included in our count. For d_1=5$d_1=5$ and d_5=0$d_5=0$, setting d_2=d_3=d_4=0$d_2=d_3=d_4=0$ gives exactly 50000, which is included in the 4608 count. textTotal numbers = 4608 - 1 = 4607$\text{Total numbers} = 4608 - 1 = 4607$ ### Pattern Recognition Always look carefully at edge constraints like "greater than". Counting the number 50000 explicitly avoids typical off-by-one errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

### Related Formula Consecutive combinations ratio property: frac^nC_r-1^nC_r = fracrn-r+1$\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{r}{n-r+1}$ ### Core Logic Setting up ratios between consecutive given coefficients: frac2856 = frac12 = fracrn-r+1 implies 3r = n + 1 quad dots (1)$\frac{28}{56} = \frac{1}{2} = \frac{r}{n-r+1} \implies 3r = n + 1 \quad \dots (1)$ frac5670 = frac45 = fracr+1n-r implies 9r = 4n - 5 quad dots (2)$\frac{56}{70} = \frac{4}{5} = \frac{r+1}{n-r} \implies 9r = 4n - 5 \quad \dots (2)$ Solving equations (1) and (2) gives r = 3$r = 3$ and n = 8$n = 8$. ### Step 1: Locating Vertices and Centroid Locus Substituting values for point C$C$ gives C(1,0)$C(1,0)$. Let the centroid coordinates be (x,y)$(x,y)$: 3x = 4cos t + 2sin t + 1 implies 3x - 1 = 4cos t + 2sin t$3x = 4\cos t + 2\sin t + 1 \implies 3x - 1 = 4\cos t + 2\sin t$ 3y = 4sin t - 2cos t + 0 implies 3y = 4sin t - 2cos t$3y = 4\sin t - 2\cos t + 0 \implies 3y = 4\sin t - 2\cos t$ ### Step 2: Squaring and Summing Trig Components Squaring and adding both parametric tracking components eliminates t$t$: (3x - 1)^2 + (3y)^2 = (4cos t + 2sin t)^2 + (4sin t - 2cos t)^2 = 16 + 4 = 20$(3x - 1)^2 + (3y)^2 = (4\cos t + 2\sin t)^2 + (4\sin t - 2\cos t)^2 = 16 + 4 = 20$ Thus, alpha = 20$\alpha = 20$. ### Pattern Recognition Symmetric parameter sets of form (Acos t + Bsin t)^2 + (Asin t - Bcos t)^2$(A\cos t + B\sin t)^2 + (A\sin t - B\cos t)^2$ collapse instantly into A^2 + B^2$A^2 + B^2$ via basic Pythagorean identities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Permutations and Combinations