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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. (\frac{n(n-1)(n-2)}{3!})
Explanation: To get combinations (3!) arrangements are removed. In exams divide ordered triples by (3!) to get unordered triples.
Correct answer: A. One object is fixed to remove rotational duplicates
Explanation: Rotations in a circle are considered the same so one position is fixed. In exams divide linear (n!) by (n) for circular permutation.
Correct answer: B. (2)
Explanation: When reflection is also same each circular arrangement is counted twice. In exams remember (\frac{(n-1)!}{2}) for necklace type.
Correct answer: A. When repetition is allowed and order is important
Explanation: Each of the (r) positions has (n) independent choices. In exams do not write a decreasing product when repetition is allowed.
Correct answer: A. Because a chosen object is not selected again
Explanation: Without repetition the available count decreases after one choice is used. In exams connect no repetition with a falling product.
Correct answer: A. Because (n!) includes arrangements of unselected objects too
Explanation: In the formula (n!) loses both unwanted tail ((n-r)!) and order (r!). In exams understand both parts of the combination denominator.
Correct answer: A. (1)
Explanation: Near the maximum (\frac{^{n}C_{r+1}}{^{n}C_r}) is around (1). In exams use the ratio to identify transition from increasing to decreasing.
Correct answer: A. (\frac{9!}{5!})
Explanation: In the falling product (9\cdot8\cdot7\cdot6), the remaining tail (5!) is removed. In exams write the denominator tail after the last chosen factor.
Correct answer: B. Divide by (4!)
Explanation: (^{10}P_4) is an ordered count and (^{10}C_4) is unordered. In exams divide by (r!) to remove order.
Correct answer: B. (^{n}C_2=\frac{^{n}P_2}{2!})
Explanation: To get unordered pairs from ordered pairs (2!) orders are removed. This connection is very useful in pair selection exams.
Correct answer: A. (\frac{n!}{p!q!})
Explanation: After choosing (p), the remaining (q) are fixed, so (\frac{n!}{p!q!}). In exams use factorial division for fixed group sizes.
Correct answer: A. Internal permutations of identical letters give the same result
Explanation: Interchanging identical letters does not create a new arrangement. In exams divide by factorials of repeated counts.
Correct answer: A. Consecutive combination ratio derivation
Explanation: It is obtained by taking the ratio of (^{n}C_r) and (^{n}C_{r-1}). In exams simplify adjacent combinations using factorial ratios.
Correct answer: A. Because (2!=2)
Explanation: The denominator in the pair formula is (2!), and (2!=2). In exams simplify pair formulas mentally.
Correct answer: A. (17!) cancels in (\frac{20!}{3!17!})
Explanation: Write (20!) as (20\cdot19\cdot18\cdot17!) and cancel (17!). In exams do not expand large factorials fully.
Correct answer: A. Complement symmetry generally does not hold in permutations
Explanation: Complement symmetry is a property of combinations, not permutations. In exams keep identities of (^{n}C_r) and (^{n}P_r) separate.
Correct answer: A. In combinations chosen and not chosen sets are complements, while in permutations ordered length changes
Explanation: Combination counts only selection so complement works. In exams do not use complement symmetry when ordered slots appear.
Correct answer: A. (^{9}C_3-^{5}C_3)
Explanation: The complement of at least (1) girl is no girl. In exams total minus unwanted is fast for at least conditions.
Correct answer: A. (^{5}C_2\cdot^{6}C_2)
Explanation: With exactly (2) women the remaining (2) must be men. In exams use cases or direct product of combinations for exactly conditions.
Correct answer: A. Treat the block as one object and use (7!\cdot2!)
Explanation: Treating two together people as one block gives (7) objects to arrange and (2!) ways inside the block. In exams use block method for together conditions.
Correct answer: A. Because any (3) points form one unique triangle and order is not important
Explanation: A triangle is determined by the set of vertices, not their order. In exams use combinations for geometric selection.
Correct answer: A. Because each object has two choices, select or not select, and the empty selection is removed
Explanation: Each object has two independent choices, so total subsets are (2^n), and for at least (1) the empty set is removed. In exams use total minus unwanted for at least conditions.
Correct answer: A. Because (x) is chosen from (r) brackets and (1) from the rest
Explanation: To form (x^r), choose (r) brackets from (n) brackets. In exams connect coefficients with selection.
Correct answer: B. Selecting or not selecting each object
Explanation: The left side adds all selection sizes and the right side gives two choices for each object. In exams remember this identity through subset counting.
Correct answer: B. Taking (k) from the first group while selecting total (r) from two groups
Explanation: In total (r) selections, the count from the first group varies as (k). In exams connect two-source selection with Vandermonde identity.
