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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. (^{n}C_r) counts selection and (^{n}P_r) counts ordered selection
Explanation: Permutation adds the arrangement of chosen objects to combination. In exams remember (P=C\times r!) conceptually.
Correct answer: C. (n-r+1)
Explanation: In the permutation recurrence, the new factor is the choices for the (r)th position, (n-r+1). In exams connect the multiplier with the next position.
Correct answer: A. (13)
Explanation: In equal combinations, different lower indices are complementary, so (5+8=n). In exams check the sum of indices in equal (C) terms.
Correct answer: B. (\frac{n-5}{6})
Explanation: The general ratio is (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}). Putting (r=6) gives (\frac{n-5}{6}).
Correct answer: C. (14)
Explanation: The ratio is (n-6+1=n-5), and (n-5=9) gives (n=14). In exams the ratio of consecutive permutations gives the last factor.
Correct answer: B. (^{14}C_{11}=\frac{^{14}P_3}{3!})
Explanation: (^{14}C_{11}=^{14}C_3), and (^{14}C_3=\frac{^{14}P_3}{3!}). In exams first use the complement to get a smaller index.
Correct answer: D. (^{10}C_7+^{10}C_6)
Explanation: Put (n=11) and (r=7) in (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}). In exams the upper index of both terms decreases by (1).
Correct answer: A. (^{n-1}C_r)
Explanation: In Pascal's identity, the other case is not choosing the special object, (^{n-1}C_r). In exams separate included and excluded cases.
Correct answer: C. Both are equal
Explanation: The coefficients are (^{10}C_3) and (^{10}C_7), and (3+7=10). In exams coefficients of complementary powers are equal.
Correct answer: A. Because (5+6=11)
Explanation: The coefficients are (^{11}C_5) and (^{11}C_6), which have complementary indices. In exams identify binomial symmetry.
Correct answer: C. (2^{n-1})
Explanation: The sum of odd indexed combinations equals the even indexed sum, (2^{n-1}). In exams remember the alternating identity from ((1-1)^n).
Correct answer: A. \(n-5\)
Explanation: The product form of a permutation is \({}^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\). Hence, its last factor is \(n-r+1\). On putting \(r=6\), we get \(n-6+1=n-5\), so option A is correct. \(n-6\) misses the required \(+1\). Exam tip: the product must contain exactly \(r\) factors; from \(n\) to \(n-5\), there are 6 factors.
Correct answer: C. (6)
Explanation: There are (r=6) positions to fill, so there are (6) factors. In exams connect the number of factors with positions.
Correct answer: D. (15)
Explanation: (^{n}C_2=\frac{n(n-1)}{2}), and (\frac{15\times14}{2}=105). In exams identify (n) using the pair formula.
Correct answer: B. (15)
Explanation: The direct answer is option B, \(n=15\). For two ordered selections, \(^{n}P_2=\frac{n!}{(n-2)!}=n(n-1)\). The given equation therefore becomes \(n(n-1)=210\). We need two consecutive numbers whose product is 210. Since \(15\times14=210\), \(n=15\). Algebraically, \(n^2-n-210=0\), which factors as \((n-15)(n+14)=0\). The possible roots are 15 and -14, but n counts objects and must be positive, so n=15. Option A, 14, gives \(14\times13=182\). Option B, 15, gives 210 and is correct. Option C, 16, gives 240. Option D, 13, gives 156. Do not divide 210 by 2: permutations count ordered pairs, so AB and BA are different.
Correct answer: C. (5!)
Explanation: (^{n}C_5=\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}). In exams keep (r!) in the denominator of (^{n}C_r).
Correct answer: C. (5)
Explanation: After cancelling ((n-5)!), there are (5) factors from (n) to (n-4). In exams the number of numerator factors is (r).
Correct answer: C. (^{10}C_5\times5)
Explanation: First there are (^{10}C_5) ways to form the committee and (5) choices for coordinator. In exams multiply when a role follows selection.
Correct answer: A. (^{10}P_5\div4!)
Explanation: One person has a distinct role and the order of the remaining (4) members is irrelevant. In exams remove overcounting of remaining members when there is a special role.
Correct answer: D. (^{7}C_3\times{}^{6}C_2)
Explanation: There is unordered selection in two categories, so combinations are multiplied. In exams count category-wise selections separately.
Correct answer: C. (^{6}C_3{}^{7}C_2+^{6}C_4{}^{7}C_1+^{6}C_5{}^{7}C_0)
Explanation: The number of girls can be (3), (4), or (5). In exams add all valid cases in at least conditions.
Correct answer: A. (^{9}C_6\times6!)
Explanation: First choose (6) letters and then arrange them in (6!) ways. In exams order is important in a word.
Correct answer: D. (9^6)
Explanation: Each position has (9) choices available again. In exams use the power rule (n^r) when repetition is allowed.
Correct answer: A. (7\times8^4)
Explanation: The first digit cannot be (0), so there are (7) choices and (8) choices for each of the remaining four places. In exams treat leading zero separately.
Correct answer: B. (7\times{}^{7}P_4)
Explanation: There are (7) non-zero choices for the first place, and then (4) places are filled from the remaining (7) digits. In exams handle the first place separately.
