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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. First choose one marked member and then choose remaining (r-1)
Explanation: Choose the marked member in (n) ways and remove overcount of (r) possible marks. In exams understand such identities by member marking.
Correct answer: B. (\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r}{n-r+1})
Explanation: Canceling factorials in consecutive combinations gives (\frac{r}{n-r+1}). In exams do not calculate full values in ratio questions.
Correct answer: A. (\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1})
Explanation: Near the largest term, the consecutive ratio changes around (1). In exams check the transition from increasing to decreasing.
Correct answer: C. ({}^{n+r-1}C_{r-1})
Explanation: In stars and bars, (n) stars and (r-1) bars are arranged. In exams use the bars method for identical distribution.
Correct answer: A. ({}^{n-1}C_{r-1})
Explanation: Give (1) ball to each box first, then distribute the remaining (n-r) balls. In exams allot the minimum first for non-empty conditions.
Correct answer: B. Complement identity
Explanation: The two lower indices sum to (n+r-1). In exams accept complementary forms in stars and bars answers.
Correct answer: B. (^{10}P_4)
Explanation: Choosing seats first and then arranging students gives (^{10}C_4\cdot4!=^{10}P_4). In exams count both positions and arrangements.
Correct answer: A. First choose the group then order that group
Explanation: Order of selected people is meaningful in a queue. In exams derive ordered selection as combination times factorial.
Correct answer: A. (\frac{8!}{3!3!2!})
Explanation: For labelled groups, group names are fixed so only internal order is removed. In exams use factorial denominators for fixed group sizes.
Correct answer: A. (2!)
Explanation: Two groups have equal size (3), so interchanging those groups gives the same distribution. In exams divide extra by factorial of equal-sized unlabelled groups.
Correct answer: C. (r^n)
Explanation: Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and distinct boxes.
Correct answer: A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n)
Explanation: Inclusion-exclusion is used to exclude empty boxes. In exams think of removing empty boxes when the word onto appears.
Correct answer: A. Because each person has (r) independent room choices
Explanation: Independent choices of distinct people multiply. In exams connect distinct-person box problems with powers.
Correct answer: B. A child may get zero coins
Explanation: When zero is allowed, place (n) stars and (r-1) bars. In exams treat ({}^{n+r-1}C_n) and ({}^{n+r-1}C_{r-1}) as the same.
Correct answer: B. ({}^{14}C_2)
Explanation: Non-negative solutions by stars and bars are ({}^{12+3-1}C_{3-1}). In exams convert equation solutions into distribution problems.
Correct answer: C. ({}^{11}C_3)
Explanation: Give (1) to each variable first, then (11) remain. In exams subtract the number of variables for positive solutions.
Correct answer: A. (^{a}C_s\cdot{}^{n-a}C_{r-s}) where there are (a) special objects
Explanation: Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.
Correct answer: B. (^{n}C_r-{}^{n-a}C_r)
Explanation: The complement of at least one special is no special. In exams total minus none is fast for at least conditions.
Correct answer: A. (^{8}C_2+{}^{8}C_4)
Explanation: Case (1): take both fixed books, case (2): take neither. In exams keep cases clear for paired restrictions.
Correct answer: A. (^{9}C_5-{}^{7}C_3)
Explanation: Subtract selections containing both particular people from total selections. In exams solve not-together selection using complement.
Correct answer: A. (8!-7!\cdot2!)
Explanation: Subtract block arrangements where (A) and (B) are together from total arrangements. In exams complement is easy for not-together permutations.
Correct answer: B. Arrange boys first and place (4) girls in (7) gaps
Explanation: Arranging boys creates (7) gaps where girls can be placed. In exams use the gap method for no two together.
Correct answer: A. (^{8}C_5\cdot5!)
Explanation: After arranging (7) consonants, (8) gaps are formed and (5) vowels are arranged. In exams choose gaps and permute vowels.
Correct answer: A. Among (n) objects, (p) of one type and (q) of another type are identical
Explanation: Internal permutations of same-type objects do not create new arrangements. In exams derive repeated-letter counts using factorial division.
Correct answer: A. Because (A) appears three times and (N) appears twice
Explanation: Interchanging identical letters gives the same word. In exams make repeated counts into denominator factorials.
