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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: A. (2(n-k-1)(n-2)!)
Explanation: There are (n-k-1) position pairs for (A,B) and (2) choices for their order. In exams count positions first in fixed-gap problems.
Correct answer: A. (2\cdot6\cdot9!)
Explanation: There are (11-4-1=6) position pairs and (A,B) can be ordered in (2) ways. In exams there is no need to separately choose the people between them.
Correct answer: A. (8!)
Explanation: After fixing (A), the position of (B) is fixed and the remaining (8) people are arranged. In exams fix one person in one-direction circular gap problems.
Correct answer: A. (8!-2\cdot7!)
Explanation: Total circular arrangements are (8!) and the adjacent block gives (2\cdot7!) ways. In exams handle circular not-adjacent by complement.
Correct answer: A. (7!\cdot8!)
Explanation: Seat the men circularly in (7!) ways and place the women in the gaps in (8!) ways. In exams do not add an extra factor (2) in circular alternation.
Correct answer: A. (9!)
Explanation: Only rotations are duplicates, so the circular count is ((10-1)!). In exams divide by (2) only after reading the reflection condition.
Correct answer: A. (\frac{9!}{2})
Explanation: In a bracelet, both rotations and reflections are considered the same. In exams use (\frac{(n-1)!}{2}) for bracelets.
Correct answer: A. (^{9}P_5)
Explanation: When the last digit is fixed as (0), there is no leading-zero issue and (5) places are filled from (9) non-zero digits. In exams separate the zero-last case.
Correct answer: A. (4\cdot8\cdot{}^{8}P_4)
Explanation: There are (4) non-zero even choices for the last digit and (8) remaining non-zero choices for the first digit. In exams handle first and last restrictions together.
Correct answer: A. (^{6}C_3\cdot4^3\cdot4^3)
Explanation: Choose the even positions and then multiply even and odd choices independently. In exams choose positions first in exactly digit-type problems.
Correct answer: A. (\sum_{i=0}^{5}(-1)^i{}^{5}C_i(5-i)^8)
Explanation: Every symbol appearing is an onto condition and missing symbols are removed. In exams solve at least once by inclusion-exclusion.
Correct answer: A. (^{6}C_4\sum_{i=0}^{4}(-1)^i{}^{4}C_i(4-i)^9)
Explanation: First choose (4) symbols and then form onto strings on them. In exams use choose set plus onto count for exactly distinct symbols.
Correct answer: A. Onto assignments from (r) positions to selected (s) symbols
Explanation: The Stirling part partitions positions into non-empty groups and (s!) assigns groups to symbols. In exams treat exactly used symbols as onto mapping.
Correct answer: A. (\frac{10!}{2!3!1!4!})
Explanation: The exponents sum to (10) and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Correct answer: A. (^{10}C_3+10\cdot9)
Explanation: The cases are: choose three (x)'s or choose one (x^2) and one (x). In exams add all cases that form the same power.
Correct answer: A. Four (x)'s, or two (x)'s and one (x^2), or two (x^2)'s
Explanation: All disjoint choices that form exponent (4) must be added. In exams make exponent partitions for polynomial coefficients.
Correct answer: A. Roots of unity filter
Explanation: The cube roots of unity filter is useful for separating modulo (3) classes. In exams identify three-step coefficient sums with an advanced filter.
Correct answer: A. (n-r>r+1)
Explanation: The ratio (\frac{n-r}{r+1}) must be greater than (1). In exams identify the increasing region by ratio.
Correct answer: A. (n-r<r+1)
Explanation: When the consecutive ratio is less than (1), the coefficients start decreasing. In exams the inequality reverses after the peak.
Correct answer: A. (2r+6=n)
Explanation: Unequal equal-combination indices are complementary. In exams set the sum of lower indices equal to the upper index.
Correct answer: A. (6)
Explanation: The complementary condition gives (3r-2+r+8=30), so (r=6). In exams solve same-index and complement cases separately.
Correct answer: A. (19)
Explanation: ({}^{n}P_5=(n-4){}^{n}P_4), so (n-4=15). In exams apply the consecutive permutation relation directly.
Correct answer: A. (5n-9r=4)
Explanation: The ratio (\frac{n-r}{r+1}=\frac{4}{5}) gives (5n-5r=4r+4). In exams cross-multiply combination ratios.
Correct answer: A. (3n-10r+3=0)
Explanation: (\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}), and cross multiplication gives the relation. In exams keep the ratio direction correct.
Correct answer: A. (^{3}C_2{}^{9}C_3+{}^{3}C_3{}^{9}C_2)
Explanation: The cases for at least (2) special are exactly (2) and exactly (3). In exams direct cases are clear for a small special group.
