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SubjectsMathematics
सूत्रों की व्युत्पत्तियाँ और उनके पारस्परिक संबंध
In Class 11 Mathematics, this topic explains how the formulas in Permutations and Combinations are derived and how they are connected. Students learn the meaning of factorial notation, develop the formulas for nPr and nCr from counting principles, and understand why nPr = r! nCr and nCr = nC(n−r). The topic also shows when to use arrangements or selections, helping students apply these relationships logically instead of relying on memorized formulas.
Correct answer: B. (6)
Explanation: Unequal equal-combination indices are complementary, so (r+r+3=15). In exams set the sum of lower indices equal to the upper index.
Correct answer: C. (5)
Explanation: The complementary condition gives (2r+r+3=18), so (r=5). In exams consider two cases for equality: same index or complementary index.
Correct answer: A. (r+s=n)
Explanation: Different indices with equal values are complementary. In exams add the indices and match them with (n).
Correct answer: A. All orders of the selected (r) objects
Explanation: Combination gives only the group and (r!) adds arrangements of that group. In exams treat permutation as selection plus arrangement.
Correct answer: A. Order is counted in permutation but not in combination
Explanation: To get unordered count from ordered count, (r!) orders are removed. In exams divide when order is ignored.
Correct answer: A. (2^n-1-n-{}^{n}C_2)
Explanation: Subtract selections of (0), (1), and (2) from all subsets. In exams handle at least by total minus small unwanted cases.
Correct answer: B. (1+n+{}^{n}C_2+{}^{n}C_3)
Explanation: At most (3) includes selections of (0), (1), (2), and (3). In exams add all allowed cases when there is an upper limit.
Correct answer: B. (^{21}C_3)
Explanation: In stars and bars, (18) stars and (3) bars are arranged. In exams use (^{n+r-1}C_{r-1}) for non-negative solutions.
Correct answer: A. (^{14}C_2)
Explanation: Giving (3) first to each variable leaves (12), then count non-negative solutions. In exams shift the minimum condition.
Correct answer: B. (^{12}C_3)
Explanation: For identical pens and distinct students, use (9) stars and (3) bars. In exams remember stars and bars for zero-allowed distribution.
Correct answer: A. (^{10}C_4)
Explanation: After giving (1) ball to each box, (6) balls remain. In exams allot the minimum first in positive distribution.
Correct answer: A. (^{4}C_1\cdot{}^{13}C_2)
Explanation: First choose the empty person, then distribute positively among the remaining (3) persons. In exams use choose empty plus positive stars and bars for exactly empty conditions.
Correct answer: A. When each object can independently go to any box
Explanation: Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and unrestricted boxes.
Correct answer: B. (10^6)
Explanation: Each prize has (10) independent choices. In exams use the power formula when distinct prizes and repeated recipients are allowed.
Correct answer: C. (^{10}P_6)
Explanation: Prizes are distinct and recipients cannot repeat, so it becomes an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Correct answer: B. (^{15}C_9)
Explanation: In identical prize distribution, use (6) stars and (9) bars. In exams apply stars and bars for identical items and distinct receivers.
Correct answer: B. (^{8}P_3)
Explanation: The three posts are different, so the order of selection is meaningful. In exams use permutation for different posts.
Correct answer: A. Because order has no importance in a committee
Explanation: A committee without posts counts only the group. In exams use combination when there are no posts.
Correct answer: B. (^{12}C_3)
Explanation: Any (3) points form a triangle and order is not important. In exams use combination in geometry selection.
Correct answer: A. (^{10}C_3-{}^{4}C_3)
Explanation: Subtract collinear (3)-point selections from total (3)-point selections. In exams subtract invalid selections.
Correct answer: A. (^{n}C_4)
Explanation: A selection of four vertices determines a quadrilateral. In exams use combinations of vertices when shapes are formed from polygons.
Correct answer: A. Because a handshake pair is unordered
Explanation: A handshake of (A) with (B) and (B) with (A) is the same. In exams treat mutual relations as combinations.
Correct answer: B. (^{n}C_2)
Explanation: A match is determined by an unordered pair of two teams. In exams use (^{n}C_2) for one-to-one pair events.
Correct answer: A. (5!\cdot3!)
Explanation: Treating the (3) special books as one block gives (5) objects to arrange. In exams also count (3!) arrangements inside the block.
Correct answer: A. (9!-8!\cdot2!)
Explanation: Subtract the arrangements where (A) and (B) are together as a block from total arrangements. In exams use complement for not together.
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