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Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to: [cite: 3383, 3384]

Solution & Explanation

### Related Formula Number of ways to select r items from n distinct items: binomnr = fracn!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].
Group grid selection diagram for Q66 - JEE Main 2025 Evening
Group grid selection diagram for Q66 - JEE Main 2025 Evening
### Step 1: Construct Mutually Exclusive Cases Let b_A, g_A represent boys and girls from Group A, and b_B, g_B from Group B [cite: 4026, 4027, 4028]. We require: b_A + b_B = 4 g_A + g_B = 4 b_A + g_A = 5 quad text(Group A total) b_B + g_B = 3 quad text(Group B total) Since Group A contains only 3 girls, g_A le 3. Since 4 boys are invited in total, b_A le 4. - **Case I:** 2 Boys & 3 Girls from Group A Rightarrow 2 Boys & 1 Girl from Group B . textWays = binom72 cdot binom33 times binom62 cdot binom51 textWays = 21 cdot 1 times 15 cdot 5 = 1575 - **Case II:** 3 Boys & 2 Girls from Group A Rightarrow 1 Boy & 2 Girls from Group B . textWays = binom73 cdot binom32 times binom61 cdot binom52 textWays = 35 cdot 3 times 6 cdot 10 = 6300 - **Case III:** 4 Boys & 1 Girl from Group A Rightarrow 0 Boys & 3 Girls from Group B . textWays = binom74 cdot binom31 times binom60 cdot binom53 textWays = 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 ### 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) to prevent generating redundant scenarios. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

Reference Study Guides

Permutations And Combinations

More Permutations and Combinations Previous-Year Questions — Page 3

Q74 2025 Numbers and Digits Sum Criteria
If the number of seven-digit numbers, such that the sum of their digits is even, is m cdot n cdot 10^n [cite: 696], where m, n in \1, 2, 3, dots, 9\ [cite: 697], then m + n is equal to[cite: 697]:
Numerical Answer. Answer: 14 to 14

Solution

### 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 [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 [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 [cite: 1468] Matching structural parameters [cite: 1469]: m = 9, quad n = 5 [cite: 1469] Both numbers belong to set range \1, 2, dots, 9\ [cite: 697]. Find target summation parameter value [cite: 1470]: 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
Q74 2025 Counting Principles
Let m and n , (m < n) be two 2-digit numbers. Then the total numbers of pairs (m, n) , such that gcd(m, n) = 6 , is
Numerical Answer. Answer: 64 to 64

Solution

### Core Logic Since gcd(m,n) = 6, we can define m = 6a and n = 6b, where a and b are coprime integers (gcd(a,b) = 1). Given that m < n, we must have a < b. Both m and n are two-digit numbers, which means 10 le m, n le 99: 10 le 6a le 99 implies 1.66 le a le 16.5 implies 2 le a le 16 10 le 6b le 99 implies 1.66 le b le 16.5 implies 2 le b le 16 Thus, we need to count all coordinate integer pairs (a,b) satisfying 2 le a < b le 16 such that gcd(a,b) = 1. ### Step 1: Systematic Counting by Fixed Value of 'a' Let's list the valid values for b for each choice of a in the range [2, 16]: - a=2: b in \3, 5, 7, 9, 11, 13, 15\ implies 7 text pairs - a=3: b in \4, 5, 7, 8, 10, 11, 13, 14, 16\ implies 9 text pairs - a=4: b in \5, 7, 9, 11, 13, 15\ implies 6 text pairs - a=5: b in \6, 7, 8, 9, 11, 12, 13, 14, 16\ implies 9 text pairs - a=6: b in \7, 11, 13\ implies 3 text pairs - a=7: b in \8, 9, 10, 11, 12, 13, 15, 16\ implies 8 text pairs - a=8: b in \9, 11, 13, 15\ implies 4 text pairs - a=9: b in \10, 11, 13, 14, 16\ implies 5 text pairs - a=10: b in \11, 13\ implies 2 text pairs - a=11: b in \12, 13, 14, 15, 16\ implies 5 text pairs - a=12: b in \13\ implies 1 text pair - a=13: b in \14, 15, 16\ implies 3 text pairs - a=14: b in \15\ implies 1 text pair - a=15: b in \16\ implies 1 text pair ### Step 2: Final Summation Summing up all valid ordered coordinate tracking entries: textTotal = 7 + 9 + 6 + 9 + 3 + 8 + 4 + 5 + 2 + 5 + 1 + 3 + 1 + 1 = 64 ### Pattern Recognition For modular subset counts, convert your boundary targets to factor conditions directly. Listing terms by prime factors reduces counting errors compared to checking every pair from scratch. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations Class 11 Mathematics: Number Theory
Q65 2025 Combinatorial Geometry
Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse fracx^216 + fracy^2n = 1 is:
  • A. frac34
  • B. frac12
  • C. fracsqrt74
  • D. frac1sqrt2

Solution

### Related Formula The combinations identity for consecutive selection values is: ^nC_r + ^nC_r+1 = ^n+1C_r+1 ### Core Logic Number of triangles from n vertices: p = ^nC_3. Number of quadrilaterals from n vertices: q = ^nC_4. Given algebraic rule: p + q = 126 implies ^nC_3 + ^nC_4 = 126 Applying Pascal's identity: ^n+1C_4 = 126 ### Step 1: Solve for n We need to find n such that ^n+1C_4 = 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 Equating the consecutive terms: n + 1 = 9 implies n = 8 ### Step 2: Calculate Eccentricity Substitute n = 8 into the ellipse equation: fracx^216 + fracy^28 = 1 Here, a^2 = 16 and b^2 = 8. e = sqrt1 - fracb^2a^2 = sqrt1 - frac816 = sqrtfrac12 = frac1sqrt2 ### 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
Q73 2025 Divisibility Principles and Counting
The number of 3-digit numbers, that are divisible by 2 and 3, but not divisible by 4 and 9, is ________.
Numerical Answer. Answer: 125

Solution

### Related Formula The number of multiples of a given integer k within a finite interval loop sequence is evaluated using standard integer division: textCount = leftlfloor fractextRange Totalk rightrfloor ### Core Logic The total count of all possible 3-digit numbers spanning from 100 to 999 is: textTotal = 999 - 100 + 1 = 900 Numbers divisible by both 2 and 3 must be multiples of their lowest common multiple, textLCM(2,3) = 6: textCount_textdiv by 6 = frac9006 = 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: textCount_textdiv by 36 = frac90036 = 25 Subtract the excluded common multiples from the initial group: textNet 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) 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

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