### 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

### Related Formula Number of ways to select r$r$ items from n$n$ distinct items: binomnr = fracn!r!(n-r)!$\binom{n}{r} = \frac{n!}{r!(n-r)!}$ ### Core Logic We need to invite a total of 4 boys and 4 girls (8 people). The constraint specifies that exactly 5 must be selected from Group A and exactly 3 from Group B. Let\'s categorize the partitions into distinct cases[cite: 4021, 4022, 4023].

### Step 1: Construct Mutually Exclusive Cases Let b_A, g_A$b_A, g_A$ represent boys and girls from Group A, and b_B, g_B$b_B, g_B$ from Group B [cite: 4026, 4027, 4028]. We require: b_A + b_B = 4$b_A + b_B = 4$ g_A + g_B = 4$g_A + g_B = 4$ b_A + g_A = 5 quad text(Group A total)$b_A + g_A = 5 \quad \text{(Group A total)}$ b_B + g_B = 3 quad text(Group B total)$b_B + g_B = 3 \quad \text{(Group B total)}$ Since Group A contains only 3 girls, g_A le 3$g_A \le 3$. Since 4 boys are invited in total, b_A le 4$b_A \le 4$. - **Case I:** 2 Boys & 3 Girls from Group A Rightarrow$\Rightarrow$ 2 Boys & 1 Girl from Group B . textWays = binom72 cdot binom33 times binom62 cdot binom51$\text{Ways} = \binom{7}{2} \cdot \binom{3}{3} \times \binom{6}{2} \cdot \binom{5}{1}$ textWays = 21 cdot 1 times 15 cdot 5 = 1575$\text{Ways} = 21 \cdot 1 \times 15 \cdot 5 = 1575$ - **Case II:** 3 Boys & 2 Girls from Group A Rightarrow$\Rightarrow$ 1 Boy & 2 Girls from Group B . textWays = binom73 cdot binom32 times binom61 cdot binom52$\text{Ways} = \binom{7}{3} \cdot \binom{3}{2} \times \binom{6}{1} \cdot \binom{5}{2}$ textWays = 35 cdot 3 times 6 cdot 10 = 6300$\text{Ways} = 35 \cdot 3 \times 6 \cdot 10 = 6300$ - **Case III:** 4 Boys & 1 Girl from Group A Rightarrow$\Rightarrow$ 0 Boys & 3 Girls from Group B . textWays = binom74 cdot binom31 times binom60 cdot binom53$\text{Ways} = \binom{7}{4} \cdot \binom{3}{1} \times \binom{6}{0} \cdot \binom{5}{3}$ textWays = 35 cdot 3 times 1 cdot 10 = 1050$\text{Ways} = 35 \cdot 3 \times 1 \cdot 10 = 1050$ ### Step 2: Total Sum Sum the combinations from all individual configurations : textTotal Ways = 1575 + 6300 + 1050 = 8925$\text{Total Ways} = 1575 + 6300 + 1050 = 8925$ ### Pattern Recognition When dealing with multi-group distributions, start your case selection using the component with the tightest constraint (here, girls in Group A le 3$\le 3$) to prevent generating redundant scenarios. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula The number of multiples of a given integer k$k$ within a finite interval loop sequence is evaluated using standard integer division: textCount = leftlfloor fractextRange Totalk rightrfloor$\text{Count} = \left\lfloor \frac{\text{Range Total}}{k} \right\rfloor$ ### Core Logic The total count of all possible 3-digit numbers spanning from 100 to 999 is: textTotal = 999 - 100 + 1 = 900$\text{Total} = 999 - 100 + 1 = 900$ Numbers divisible by both 2 and 3 must be multiples of their lowest common multiple, textLCM(2,3) = 6$\text{LCM}(2,3) = 6$: textCount_textdiv by 6 = frac9006 = 150$\text{Count}_{\text{\div by 6}} = \frac{900}{6} = 150$ ### Step 1: Apply Set Inclusion-Exclusion for Constraints The problem asks to exclude numbers divisible by 4 and 9. This means we must remove any number that is a multiple of 6 and also a multiple of textLCM(4,9) = 36$\text{LCM}(4,9) = 36$: textCount_textdiv by 36 = frac90036 = 25$\text{Count}_{\text{\div by 36}} = \frac{900}{36} = 25$ Subtract the excluded common multiples from the initial group: textNet Count = 150 - 25 = 125$\text{Net Count} = 150 - 25 = 125$ ### Pattern Recognition Phrases like 'divisible by A and B but not by C and D' can be simplified using set theory concepts by analyzing the lowest common multiples (textLCM$\text{LCM}$) of the underlying divisibility rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Principle of Mathematical Induction