Let S = \(m, n): m, n in \1, 2, 3, dots, 50\ \ . If the number of elements (m, n) in S such that 6^m + 9^n is a multiple of 5 is p and the number of elements (m, n) in S such that m + n is a square of a prime number is q , then p + q is equal to......

Numerical Answer Type:
Enter a numerical value Answer: 1333 to 1333 +4 marks

Solution & Explanation

### Related Formula Modular arithmetic reductions for power cycles: a equiv b pmodm Rightarrow a^k equiv b^k pmodm ### Core Logic Analyze condition p: (6^m + 9^n) is divisible by 5. 6 equiv 1 pmod 5 Rightarrow 6^m equiv 1^m equiv 1 pmod 5. 9 equiv -1 pmod 5 Rightarrow 9^n equiv (-1)^n pmod 5. For the sum to be divisible by 5: 1 + (-1)^n equiv 0 pmod 5 Rightarrow (-1)^n = -1. This implies n must be an ODD integer. Since m in \1, 2, dots, 50\, m can be anything (50 choices). Since n must be odd in \1, dots, 50\, n has 25 choices. p = 50 times 25 = 1250. ### Step 1: Compute q Analyze condition q: (m + n) is the square of a prime number. Max value of m+n = 50+50 = 100. Primes whose squares are leq 100: 2, 3, 5, 7. Their squares are 4, 9, 25, 49. So m+n can be 4, 9, 25, 49. Match List-I with List-II:
m+n=4m+n=9m+n=25m+n=49
No. of ways382448
Explanation for counts: If m+n = S, and m, n geq 1, the number of ways is S-1 (since S leq 50). For S=4: 3 ways. For S=9: 8 ways. For S=25: 24 ways. For S=49: 48 ways. q = 3 + 8 + 24 + 48 = 83. ### Step 2: Final Sum p + q = 1250 + 83 = 1333 ### Pattern Recognition Modular exponentiation immediately shrinks large powers to pm 1. The sum m+n=S where 1 le m,n le N has exactly S-1 solutions if S le N, allowing instant combinatorics tallying without manual counting. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

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Permutations And Combinations

More Permutations and Combinations Previous-Year Questions — Page 2

Q58 jee_main_2025_03_april_evening Geometry Problems
Line L_1 of slope 2 and line L_2 of slope frac12 intersect at the origin O. In the first quadrant, P_1, P_2, dots, P_12 are 12 points on line L_1 and Q_1, Q_2, dots, Q_9 are 9 points on line L_2. Then the total number of triangles, that can be formed having vertices at three of the 22 points O, P_1, P_2, dots, P_12, Q_1, Q_2, dots, Q_9, is:
  • A. 1080
  • B. 1134
  • C. 1026
  • D. 1188

Solution

### Related Formula Number of ways to choose 3 points out of N points is ^N C_3. If any points are collinear, choosing 3 points from those collinear points will form a straight line instead of a triangle. ### Core Logic Total number of points = 1 (origin O) + 12 (on L_1) + 9 (on L_2) = 22 points. Triangles are formed by choosing any 3 points except those that are collinear. ### Step 1: Identifying Collinear Sets 1. The set of points lying on L_1 includes O, P_1, dots, P_12, which is 13 collinear points. Number of collinear combinations = ^13 C_3 2. The set of points lying on L_2 includes O, Q_1, dots, Q_9, which is 10 collinear points. Number of collinear combinations = ^10 C_3 ### Step 2: Triangle Calculation Total Triangles = Total choices of 3 points - Collinear choices on L_1 - Collinear choices on L_2 textTotal = ^22 C_3 - ^13 C_3 - ^10 C_3 Calculating individual combinations: - ^22 C_3 = frac22 times 21 times 203 times 2 times 1 = 1540 - ^13 C_3 = frac13 times 12 times 113 times 2 times 1 = 286 - ^10 C_3 = frac10 times 9 times 83 times 2 times 1 = 120 textTriangles = 1540 - 286 - 120 = 1134 ### Pattern Recognition Alternatively, count using partition combinations to avoid large factorials: textTriangles = (^12 C_2 times ^9 C_1) + (^9 C_2 times ^12 C_1) + (1 times ^12 C_1 times ^9 C_1) = (66 times 9) + (36 times 12) + (108) = 594 + 432 + 108 = 1134 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations
Q57 jee_main_2025_07_april_morning Practical Problems on Combinations
From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is
  • A. 165
  • B. 155
  • C. 145
  • D. 135

Solution

### Related Formula Number of ways to select r items from a pool of n distinct objects: ^nC_r = fracn!r!(n-r)! ### Core Logic Total required players = 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 players. - Available remaining pool: - Batsmen available = 7 - 1 = 6 batsmen. - Bowlers available = 6 - 1 = 5 bowlers. Adjusted structural constraints for the remaining 8 slots: - Needs at least 4 - 1 = 3 more batsmen. - Needs at least 4 - 1 = 3 more bowlers. ### Step 1: Set Up Case Combinations Let x be the number of additional batsmen and y be the number of additional bowlers selected, where x + y = 8 with x ge 3 and y ge 3. Possible case matches: - **Case 1**: 5 batsmen and 3 bowlers (x=5, y=3) - **Case 2**: 4 batsmen and 4 bowlers (x=4, y=4) - **Case 3**: 3 batsmen and 5 bowlers (x=3, y=5) ### Step 2: Calculate Each Case Value 1. For Case 1: textWays = ^6C_5 times ^5C_3 = 6 times 10 = 60 2. For Case 2: textWays = ^6C_4 times ^5C_4 = 15 times 5 = 75 3. For Case 3: textWays = ^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 ### 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) and the final destination requirement count (r) before designing your target case distributions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations
Q75 jee_main_2025_07_april_morning Subsets without Consecutive Elements
For n geq 2 , let S_n denote the set of all subsets of \1, 2, dots, n\ with no two consecutive numbers. For example \1, 3, 5\ in mathbfS_6 , but \1, 2, 4\ notin mathbfS_6 . Then n(mathbfS_5) is equal to
Numerical Answer. Answer: 13 to 13

Solution

### Related Formula The number of ways to choose r non-consecutive elements from a set of n items is given by the binomial formula: textWays = ^n - r + 1C_r ### Core Logic We need to find the total number of subsets for n = 5 items without consecutive values. We break down the options based on the size of the subset (r), ranging from an empty set (r=0) up to the maximum possible non-consecutive size (r=3). ### Step 1: Compute Combinations for Each Size 1. Subsets containing **no elements** (r = 0): textWays = 1 quad (textThe empty set emptyset) 2. Subsets containing **exactly 1 element** (r = 1): textWays = ^5 - 1 + 1C_1 = ^5C_1 = 5 quad (\1\, \2\, \3\, \4\, \5\) 3. Subsets containing **exactly 2 elements** (r = 2): textWays = ^5 - 2 + 1C_2 = ^4C_2 = 6 4. Subsets containing **exactly 3 elements** (r = 3): textWays = ^5 - 3 + 1C_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 ### Pattern Recognition Shortcut: The total number of non-consecutive subsets for a set of size n follows the Fibonacci sequence pattern: F_n+2. For n=5, the value is the 7th Fibonacci number, which is exactly 13. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations
Q63 jee_main_2025_08_april_evening Geometry Problems in Combinatorics
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
  • A. 230
  • B. 220
  • C. 200
  • D. 210

Solution

### Related Formula textNet Triangles = binomn3 - binomm3 ### 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 ### Step 2: Isolate Internal Flattened Collinear Triplets Choosing 3 items completely bundled inside the collinear group row of size 5: binom53 = 10 ### Step 3: Deduce Triangles textTriangles 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
Q60 jee_main_2025_29_jan_evening Rank of a Word in Dictionary
If all the words with or without meaning made using all the letters of the word “KANPUR” are arranged as in a dictionary, then the word at 440^textth position in this arrangement, is :
  • A. textPRNAKU
  • B. textPRKANU
  • C. textPRKAUN
  • D. textPRNAUK

Solution

### Related Formula Number of permutations of n distinct letters is given by: textPermutations = n! ### Core Logic Sort the distinct letters of "KANPUR" alphabetically: textA, K, N, P, R, U Track alphabetical structural sets row by row: Words starting with A: 5! = 120 Words starting with K: 5! = 120 (Cumulative: 240) Words starting with N: 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 (Cumulative: 384) Words starting with PK: 4! = 24 (Cumulative: 408) Words starting with PN: 4! = 24 (Cumulative: 432) ### Step 2: Reach the Targeted Count Remaining difference to hit 440 is 440 - 432 = 8 words. The next group starts with PR. Alphabetical listings within PR: Words starting with PRA: 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

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