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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: C. Choosing the special object and then choosing remaining (r-1)
Explanation: If the special object is chosen the remaining (r-1) objects must be chosen from (n-1). In exams write the included case separately.
Correct answer: D. (^{5}C_2+^{5}C_1)
Explanation: Put (n=6) and (r=2) in (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}). In exams (n) decreases by (1) in both terms.
Correct answer: B. Each object has two choices take or not take
Explanation: Each object has two choices so total subsets are (2^n). In exams think of subsets using include or exclude.
Correct answer: D. Because the empty selection is removed from total (2^n) selections
Explanation: All selections include the empty selection too. In exams subtract the empty case when at least one appears.
Correct answer: C. (^{n}C_r)
Explanation: The direct answer is option C, \(^{n}C_r\). A subset is a group selected from a larger set, and its order does not matter: choosing A, B is the same subset as choosing B, A. The notation \(^{n}C_r=\frac{n!}{r!(n-r)!}\) counts the ways to select r distinct objects from n objects without arranging them. Option A, \(^{n}P_r\), counts ordered arrangements, so it counts the same group repeatedly in different orders. Option B, \(n!\), arranges all n objects and is not restricted to selecting r. Option D, \(r!\), only counts arrangements of the selected objects and does not choose them from n. Thus C is the required subset formula. Exam cue: subset means unordered selection, so think combination.
Correct answer: B. (^{4}C_2\times{}^{5}C_1)
Explanation: Two independent selections are needed together so the multiplication principle applies. In exams the word and often means multiply.
Correct answer: A. (^{8}C_3\times3!)
Explanation: First choose (3) students and then arrange them in (3!) ways. In exams multiply when arrangement follows selection.
Correct answer: D. ((n-1)!)
Explanation: Rotations are considered same in a circle so fixing one object gives ((n-1)!) arrangements. In exams fix (1) object in circular arrangement.
Correct answer: A. Because each circular arrangement is counted in (n) rotations
Explanation: Rotating a circle does not create a new arrangement so (n!) is divided by (n). In exams treat rotation as duplicate counting.
Correct answer: A. Because internal orders of identical letters are not different
Explanation: Interchanging identical letters does not create a new word. In exams divide by the factorial of identical letters.
Correct answer: B. When two types of objects are identical (p) and (q) times respectively
Explanation: Interchanges inside two identical groups are not counted separately so we divide by (p!q!). In exams place the factorial of each repeated group in the denominator.
Correct answer: C. (n^r)
Explanation: Each position independently has (n) choices so the product is (n\times n\times\cdots\times n=n^r). In exams choices do not decrease when repetition is allowed.
Correct answer: A. Because choices decrease at each next position
Explanation: Without repetition one object cannot be taken again. In exams pay attention to allowed and not allowed words.
Correct answer: C. Count ordered arrangements first then remove extra (r!) orders
Explanation: Combination is obtained by removing (r!) internal orders from permutation. In exams make the ordered count first and then divide.
Correct answer: A. (\frac{n!}{(n-r+1)!})
Explanation: The direct answer is option A, \(\frac{n!}{(n-r+1)!}\). Begin with the given rule \(^{n}P_k=\frac{n!}{(n-k)!}\). To find \(^{n}P_{r-1}\), replace every occurrence of \(k") with \(r-1\): \(^{n}P_{r-1}=\frac{n!}{(n-(r-1))!}\). Simplify the denominator: \(n-(r-1)=n-r+1\). Therefore \(^{n}P_{r-1}=\frac{n!}{(n-r+1)!}\). Option A has exactly this denominator and is correct. Option B, \(\frac{n!}{(n-r-1)!}\), comes from changing the sign incorrectly. Option C reverses numerator and denominator and is not the permutation formula. Option D uses \(r!\), which is unrelated to the required substitution. The safe exam method is to put parentheses around \(r-1\) before subtracting: \(n-(r-1)\), then simplify carefully.
Correct answer: B. (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r})
Explanation: Simplifying the factorial formula gives the ratio (\frac{n-r+1}{r}). In exams ratio method is useful for adjacent combinations.
Correct answer: B. (^{n}P_r=^{n}P_{r-1}\times(n-r+1))
Explanation: For the (r)th position (n-r+1) choices are added as a factor. In exams identify the factor of the next position.
Correct answer: A. Double counting by marking one chosen object
Explanation: In each (r)-selection one of the (r) chosen objects can be marked. In exams understand this relation through marked object counting.
Correct answer: B. From adjacent combination ratio
Explanation: It is obtained by simplifying the ratio of (^{n}C_r) and (^{n}C_{r-1}). In exams use ratio for consecutive (C) terms.
Correct answer: C. (^{n}C_r\times r!)
Explanation: There are (^{n}C_r) ways to choose (r) objects and then (r!) arrangements. In exams write ordered selection as choose and arrange.
Correct answer: B. (^{7}C_3)
Explanation: Choosing (4) means leaving (3) so (^{7}C_4=^{7}C_3). In exams look at both chosen and not chosen.
Correct answer: C. (10)
Explanation: The direct answer is option C, 10. For a permutation, the first position has 12 choices, the second has 11 choices because one object has already been used, and the third has 10 choices because two objects have been used. Thus, \(^{12}P_3=12\times11\times10\), so the last factor is 10. Option A, 12, is the first factor, not the last. Option B, 11, is the second factor. Option C, 10, is correct because it equals \(12-3+1\). Option D, 9, would be one factor too far; only three factors are required. The useful general pattern is \(^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\). Memory cue: for three selections, count down from \(n\) three times; the last number is \(n-2\).
Correct answer: A. 11C3
Explanation: The governing identity is the symmetry property of combinations: nCr = nC(n-r). Selecting r objects from n is equivalent to deciding which n-r objects are left out, so both selections have the same count. Applying this to 11C8 gives 11C8 = 11C(11-8) = 11C3. The smaller lower index makes calculation easier, and 11C3 can be evaluated as 11 × 10 × 9 / (3 × 2 × 1) = 165. Thus option A is correct. Option B changes the total number from 11 to 8, so it is not equivalent. Option C is a permutation and counts ordered arrangements, while option D asks for 11 selections from only 3 objects and is not valid in the usual combinatorial setting.
Correct answer: C. (^{6}P_6=6!)
Explanation: The row arrangement of all (6) distinct books is (^{6}P_6=6!). In exams connect all distinct arrangement with factorial.
Correct answer: B. (^{5}C_3\times3!)
Explanation: Order is important when placing from top to bottom so the chosen (3) colours are arranged in (3!) ways. In exams add arrangement when positions appear.
