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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. (^{n}P_r=^{n}C_r\times r!)
Explanation: Each selected group can be arranged in (r!) ways. In exams treat permutation as selection followed by arrangement.
Correct answer: A. Because arrangements of (r) objects are not counted
Explanation: In combinations order is not important so (r!) duplicate arrangements are removed. In exams divide when order is ignored.
Correct answer: B. (^{n}P_r=\frac{n!}{(n-r)!})
Explanation: After (r) decreasing factors the remaining tail is ((n-r)!). In exams put the missing tail in the denominator.
Correct answer: B. Choosing (r) and leaving (n-r) are equivalent decisions
Explanation: Choosing (r) objects is equivalent to not choosing (n-r) objects. In exams use complement selection to identify symmetry.
Correct answer: A. A fixed object is included or not included
Explanation: Make cases on one fixed object: include it or exclude it. In exams derive such identities by inclusion cases.
Correct answer: B. (\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1})
Explanation: Writing factorial forms and canceling common terms gives the ratio. In exams ratio method is fast for consecutive combinations.
Correct answer: B. (^{n}C_r=\frac{^{n}P_r}{r!})
Explanation: A permutation includes (r!) orders for each combination. In exams divide ordered count by (r!) to get unordered count.
Correct answer: B. Because (E) appears twice
Explanation: Interchanging identical (E)'s does not give a new arrangement. In exams divide by factorials of repeated identical items.
Correct answer: B. (^{7}P_3)
Explanation: Both selection and order are needed so (^{7}P_3=^{7}C_3\cdot3!). In exams use permutation when order matters.
Correct answer: A. Because the two posts are different
Explanation: Switching captain and vice-captain changes the outcome. In exams treat different roles as order important.
Correct answer: C. Because only a group is needed
Explanation: Only membership is counted in a committee, not order. In exams use combination for a group without posts.
Correct answer: B. Choices are (n), then (n-1), then decreasing
Explanation: After each chosen object available choices decrease. In exams write decreasing products for distinct arrangements.
Correct answer: C. (^{n}C_r=^{n}C_{n-r})
Explanation: This is the symmetry of choosing (r) and leaving (n-r). In exams use this relation when opposite indices sum to (n).
Correct answer: B. \(2!\)
Explanation: The relation between permutations and combinations is \(^{n}P_r=^{n}C_r\cdot r!\). Here \(r=2\), so \(^{n}P_2=^{n}C_2\cdot 2!\), giving \(k=2!\). Note that \(2\) is numerically equal to \(2!\), so listing both as separate options would create two correct answers; always check equivalent expressions in MCQs. Exam tip: order matters in permutations, so each combination is multiplied by \(r!\).
Correct answer: A. There is one way to choose no object
Explanation: The empty selection is also a valid selection. In exams remember both (0!) and empty choice as (1).
Correct answer: A. There is exactly one way to choose all (n) objects
Explanation: When all objects are chosen the selection is fixed. In exams quickly identify extreme cases (^{n}C_0) and (^{n}C_n).
Correct answer: B. Choosing r objects from n distinct objects and arranging them in r distinct positions
Explanation: First choose r objects in \(\binom{n}{r}\) ways, then arrange them in \(r!\) orders. Thus it represents selection followed by ordering; a committee has no order. Exam tip: multiply by \(r!\) whenever positions or order matter.
Correct answer: A. \(4!\)
Explanation: The relation is \(^{n}P_r=^{n}C_r\cdot r!\), because the \(r\) selected objects can be arranged in \(r!\) orders. Here \(r=4\), so \(x=4!\). \(\frac{12!}{8!}\) is \(^{12}P_4\) itself, while \(\frac{12!}{4!8!}\) is \(^{12}C_4\) itself. Exam tip: in the relation between permutations and combinations, the multiplying factor is always \(r!\).
Correct answer: B. (^{6}P_4=^{6}C_4\cdot4!)
Explanation: Changing order changes the code. In exams usually use permutation relation for codes or passwords.
Correct answer: A. There are (n) independent choices for selecting one object
Explanation: In single selection order does not arise and choices are (n). In exams both (^{n}C_1) and (^{n}P_1) equal (n).
Correct answer: A. Because the order of (1) object creates no different result
Explanation: One object can be arranged in (1!) way. In exams permutation and combination give the same value for (r=1).
Correct answer: A. Because ordered pairs are divided by (2!) to become unordered pairs
Explanation: (AB) and (BA) are the same pair so repetition is removed. In exams (\frac{n(n-1)}{2}) is often useful for pairs.
Correct answer: B. Pascal identity
Explanation: This is a shifted form of Pascal identity. In exams add adjacent lower-row combinations to form the upper-row combination.
Correct answer: B. (5)
Explanation: The direct answer is option B: r = 5. The key idea is the symmetry property of combinations: for a fixed n, \\(^{n}C_k=^{n}C_{n-k}\\). Thus, two combination values with the same upper number are equal when their lower numbers are complementary, so r+(r+4)=14. Now solve step by step: r+r+4=14; 2r+4=14; 2r=10; r=5. Therefore, \\(^{14}C_5=^{14}C_9\\), and r+4=9, confirming the result. Option A, r=4, would give lower indices 4 and 8; these are not complementary because 4+8=12, not 14, so the values are not equal. Option B, r=5, gives 5 and 9, whose sum is 14, so it is correct. Option C, r=6, gives 6 and 10; their sum is 16, so it does not satisfy the symmetry condition. Option D, r=7, gives 7 and 11; their sum is 18, so it is also wrong. Remember: when \\(^{n}C_a=^{n}C_b\\) in this standard situation, use a+b=n, then solve carefully.
Correct answer: A. (3!)
Explanation: First choose (3) objects and arrange them in (3!) ways. In exams remember (^{n}P_3=^{n}C_3\cdot3!).
