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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. (D_5=44)
Explanation: This is the derangement of (5) objects and (D_5=44). In exams treat letters-envelope mismatch as a derangement pattern.
Correct answer: A. (^{n}C_kD_{n-k})
Explanation: First choose the (k) correctly seated people, then derange the remaining (n-k). In exams use choose fixed plus derange rest for exactly fixed points.
Correct answer: A. Each circular arrangement is counted as (n) rotations in linear arrangements
Explanation: Rotations are duplicates in the linear count (n!). In exams divide rotational overcount by (n) in circular arrangements.
Correct answer: A. When both rotations and reflections are considered the same
Explanation: In a bracelet, mirror images are considered the same. In exams divide the circular count by (2) when reflection is the same.
Correct answer: A. (5!\cdot3!)
Explanation: The block of three people and the remaining (5) people form (6) circular objects, giving (5!) arrangements. In exams use one less factorial for circular block objects.
Correct answer: A. ((n-1)!-2(n-2)!)
Explanation: Subtract adjacent block arrangements of (A,B) from total circular arrangements. In exams handle circular not-adjacent by complement.
Correct answer: A. (n!(n-1)!)
Explanation: First seat men in a circle in ((n-1)!) ways, then place women in gaps in (n!) ways. In exams fix one group first for circular alternation.
Correct answer: A. (\frac{n!}{k!})
Explanation: Only (1) of the (k!) relative orders of the special objects is allowed. In exams divide total arrangements by (k!) for fixed relative order.
Correct answer: B. (\frac{10!}{4})
Explanation: Each of the two independent relative-order restrictions halves the count. In exams divide by (2^k) for independent before-after pairs.
Correct answer: A. Because (0) is valid at the last place but invalid at the first place
Explanation: In an even number, (0) may be the unit digit but cannot be the leading digit. In exams keep zero cases separate in digit problems.
Correct answer: A. (6\cdot5\cdot4+5\cdot5\cdot4)
Explanation: The last digit can be (0) or (5), and (0) changes the first-digit restriction. In exams make unit-digit cases for divisibility by (5).
Correct answer: A. (^{4}C_2\cdot3^2\cdot4^2)
Explanation: Choose the even positions, then each has (3) even choices and the rest have (4) odd choices. In exams choose positions first in exactly conditions.
Correct answer: A. (n^r-{}^{n}P_r)
Explanation: Subtract all-distinct strings from total repetition-allowed strings. In exams solve at least repeat by the no-repeat complement.
Correct answer: A. (^{n}C_s s! S(r,s))
Explanation: First choose (s) symbols, then map the (r) positions onto those (s) symbols. In exams use the onto idea for exactly distinct symbols.
Correct answer: A. (\frac{n!}{p!q!r!})
Explanation: It is the multinomial count of choosing (a) from (p) brackets, (b) from (q), and (c) from (r). In exams treat multinomial coefficients like repeated arrangements.
Correct answer: A. (\frac{8!}{3!2!3!})
Explanation: The exponents sum to (8), and the coefficient comes from the multinomial form. In exams treat powers as group sizes.
Correct answer: A. By adding ((1+1)^n) and ((1-1)^n)
Explanation: On adding, odd-index terms cancel out. In exams use (x=1) and (x=-1) together for even-odd binomial sums.
Correct answer: A. Roots of unity filter
Explanation: The roots of unity filter separates indices modulo (3). In exams recognize that such sums differ from the usual even-odd method.
Correct answer: A. (\frac{{}^{n}C_{r-1}}{{}^{n}C_{r-2}}=\frac{n-r+2}{r-1})
Explanation: Factorial cancellation in consecutive combinations gives this ratio. In exams use adjacent ratios instead of calculating long values.
Correct answer: A. The sequence increases when the ratio is greater than (1) and decreases when it is less than (1)
Explanation: Near the maximum, the sequence transitions from increasing to decreasing. In exams locate the binomial coefficient peak by ratios.
Correct answer: A. (r=\frac{n}{2})
Explanation: For even (n), the middle index is single. In exams identify the central term by symmetry and ratio.
Correct answer: A. ({}^{n}C_{\frac{n-1}{2}}) and ({}^{n}C_{\frac{n+1}{2}})
Explanation: For odd (n), the two central complementary indices give equal maxima. In exams remember two middle terms in the odd case.
Correct answer: B. (4)
Explanation: Unequal equal-combination indices are complementary, so (3r-1+r+5=20). In exams set the sum of lower indices equal to the upper index.
Correct answer: C. (12)
Explanation: ({}^{n}P_3=(n-2){}^{n}P_2), so (n-2=10). In exams solve quickly using consecutive permutation relations.
Correct answer: A. (3n-5r=2)
Explanation: The ratio (\frac{n-r}{r+1}=\frac{2}{3}) gives (3n-3r=2r+2). In exams cross-multiply consecutive combination ratios.
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