Let m$m$ and n$n$ , (m < n)$(m < n)$ be two 2-digit numbers. Then the total numbers of pairs (m, n)$(m, n)$ , such that gcd(m, n) = 6$\gcd(m, n) = 6$ , is

### 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

### Related Formula Parity property for numerical spaces: Across any continuous span sequence ending in zero, exactly half of the configuration combinations form an even sum of digits while the rest form odd outputs. ### Core Logic Calculate the total possible combinations of 7-digit numbers first [cite: 1465]: textTotal Numbers = 9 times 10 times 10 times 10 times 10 times 10 times 10 = 9,000,000$$\text{Total Numbers} = 9 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 9,000,000$$ [cite: 1465] By using fundamental parity distributions across digit configurations, exactly half of these total options have an even sum of digits [cite: 1467]: textEven Sum Count = frac9,000,0002 = 4,500,000$$\text{Even Sum Count} = \frac{9,000,000}{2} = 4,500,000$$ [cite: 1467] ### Step 1: Extracting factors Express the final numeric amount in requested base-10 exponential format shape [cite: 1468]: 4,500,000 = 45 times 10^5 = 9 cdot 5 cdot 10^5$$4,500,000 = 45 \times 10^5 = 9 \cdot 5 \cdot 10^5$$ [cite: 1468] Matching structural parameters [cite: 1469]: m = 9, quad n = 5$$m = 9, \quad n = 5$$ [cite: 1469] Both numbers belong to set range \1, 2, dots, 9\$\{1, 2, \dots, 9\}$ [cite: 697]. Find target summation parameter value [cite: 1470]: m + n = 9 + 5 = 14$$m + n = 9 + 5 = 14$$ [cite: 1470] ### Pattern Recognition The last slot digit completely decides final parity choices. This allows splitting total permutation blocks directly in half without tedious calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula The combinations identity for consecutive selection values is: ^nC_r + ^nC_r+1 = ^n+1C_r+1$$^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$$ ### Core Logic Number of triangles from n$n$ vertices: p = ^nC_3$p = ^nC_3$. Number of quadrilaterals from n$n$ vertices: q = ^nC_4$q = ^nC_4$. Given algebraic rule: p + q = 126 implies ^nC_3 + ^nC_4 = 126$$p + q = 126 \implies ^nC_3 + ^nC_4 = 126$$ Applying Pascal's identity: ^n+1C_4 = 126$^{n+1}C_4 = 126$ ### Step 1: Solve for n We need to find n$n$ such that ^n+1C_4 = 126$^{n+1}C_4 = 126$: frac(n+1)n(n-1)(n-2)24 = 126$$\frac{(n+1)n(n-1)(n-2)}{24} = 126$$ (n+1)n(n-1)(n-2) = 3024 = 9 cdot 8 cdot 7 cdot 6$$(n+1)n(n-1)(n-2) = 3024 = 9 \cdot 8 \cdot 7 \cdot 6$$ Equating the consecutive terms: n + 1 = 9 implies n = 8$$n + 1 = 9 \implies n = 8$$ ### Step 2: Calculate Eccentricity Substitute n = 8$n = 8$ into the ellipse equation: fracx^216 + fracy^28 = 1$$\frac{x^2}{16} + \frac{y^2}{8} = 1$$ Here, a^2 = 16$a^2 = 16$ and b^2 = 8$b^2 = 8$. e = sqrt1 - fracb^2a^2 = sqrt1 - frac816 = sqrtfrac12 = frac1sqrt2$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$ ### Pattern Recognition Pascal's combination identity avoids dealing with tedious polynomial expansions when solving multi-vertex geometry systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Conic Sections