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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. E = {0, 1, 2, 3}
Explanation: The governing concept is translating a condition into an explicit roster. Decimal digits are 0 through 9, and the strict inequality x < 4 selects 0, 1, 2, and 3. The number 4 is excluded because the condition is not x ≤ 4. Therefore option A is correct. B and C wrongly include 4, while D wrongly omits the digit 0.
Correct answer: A. There are (n) choices for the first position
Explanation: Any one of the (n) distinct objects can occupy the first position. In exams think by filling positions one by one.
Correct answer: B. (r)
Explanation: Since (r) positions are to be filled there are (r) factors. In exams connect the number of factors with the number of positions.
Correct answer: B. (^{n}P_r=\frac{n!}{(n-r)!})
Explanation: In ordered selection the remaining ((n-r)!) part is removed from (n!). In exams remember ((n-r)!) in the denominator for permutation.
Correct answer: D. Because order has to be removed
Explanation: In combination order is not counted so (r!) orders of selected objects are removed. In exams treat combination as selection without order.
Correct answer: C. (^{n}P_r=^{n}C_r\times r!)
Explanation: First (r) objects are chosen and then arranged in (r!) ways. In exams keep the idea of choose first and arrange later.
Correct answer: B. Dividing (^{n}P_r) by (r!)
Explanation: In combination extra orders are removed from permutation. In exams write (^{n}P_r) and divide by (r!).
Correct answer: D. Choosing (r) objects is same as leaving (n-r) objects
Explanation: When (r) are chosen the (n-r) objects are automatically left. In exams remember the connection between choosing and leaving.
Correct answer: C. (7)
Explanation: By (^{n}C_r=^{n}C_{n-r}) we get (k=10-3=7). In exams check the complementary index in equal combinations.
Correct answer: A. Because there is only (1) way to choose no object
Explanation: Selecting zero objects is an empty selection and it happens in one way. In exams count empty selection as one way.
Correct answer: D. There is only (1) way to choose all objects
Explanation: The direct answer is option D: there is only one way to choose all n objects. A combination counts selections, not arrangements. If we must choose n objects from a collection containing exactly n objects, every object must be included. There is no freedom of choice, so the result is one selection, written as \(^{n}C_n=1\). Option A is wrong because choosing no object describes \(^{n}C_0=1\), not this expression. Option B is wrong because changing order belongs to permutations; combinations ignore order. Option C is wrong because the objects need not be identical; they are normally distinct. Option D is correct because the complete set itself is the single possible subset. Remember: choosing none or choosing all gives exactly one combination.
Correct answer: D. One empty way to arrange zero objects
Explanation: The direct answer is option D: one empty way to arrange zero objects. A permutation counts an ordered arrangement of selected objects. If zero objects are selected, there are no positions to fill, and there is exactly one possibility: do nothing, producing the empty arrangement. Algebraically, \({}^{n}P_0=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1\), and this agrees with the convention \(0!=1\). Option A, complete arrangement of \(n\) objects, describes \({}^{n}P_n=n!\), not \({}^{n}P_0\). Option B is false because \(0!\) is valid and equals 1. Option C is false: an empty arrangement is allowed when zero places are being filled. Option D correctly expresses the one empty choice.
Correct answer: D. To keep (^{n}C_n=\frac{n!}{n!0!}=1) we need (0!=1)
Explanation: The direct answer is option D. The combination formula is \(^{n}C_r=\frac{n!}{r!(n-r)!}\). For \(r=n\), it becomes \(^{n}C_n=\frac{n!}{n!0!}\). Choosing all \(n\) objects has exactly one way, so this expression must equal 1. Since \(n!/n!=1\), the denominator must contain \(0!=1\), giving \(\frac{1}{0!}=1\). Thus the convention \(0!=1\) keeps the boundary case of the combination formula correct. Option A is not the best statement: although \(^{n}C_n=1\), it does not equal \(n!\) in general, so its wording is false. Option B is wrong because \(^{n}P_n=n!\), not 0. Option C is wrong because \(^{n}C_0=1\), not 0. Option D states the exact correct connection. A useful memory cue is that choosing none or choosing everything has one way; the factorial formula must therefore use \(0!=1\).
Correct answer: A. Because (n) distinct objects are arranged in all (n) places
Explanation: Arranging all objects gives (n(n-1)\cdots1=n!). In exams connect complete arrangement with factorial.
Correct answer: B. (5\times4)
Explanation: There are (5) choices for the first place and (4) for the second. In exams choices decrease when repetition is not allowed.
Correct answer: D. (^{6}P_3)
Explanation: Seating in positions depends on order so permutation is used. In exams when position seat or rank appears check order.
Correct answer: C. (^{6}C_3)
Explanation: A team has no order so combination is used. In exams remove order for words like group or committee.
Correct answer: D. (^{7}P_2=^{7}C_2\times2!)
Explanation: First choose (2) objects and then arrange them in (2!) ways. In exams understand permutation as combination times arrangement.
Correct answer: B. (^{8}C_2=\frac{^{8}P_2}{2!})
Explanation: To get combination ordered arrangements are divided by (2!). In exams divide to remove internal orders.
Correct answer: C. nC1 = n
Explanation: The combination nCr counts selections of r objects from n objects when order does not matter. For r = 1, we are simply choosing one object from n available objects, so there are exactly n possible selections. Algebraically, nC1 = n!/[1!(n−1)!] = n!/(n−1)! = n. Therefore option C is correct. It is not 1, because there are n different single-object choices; it is not n!, because factorial notation counts far more arrangements than required; and it is not 0, because a one-object selection is possible whenever n is at least 1. This identity is also the simplest boundary case of the combination formula.
Correct answer: A. (^{n}P_1=n)
Explanation: Only one position is to be filled so there are (n) choices. In exams for (r=1) permutation and combination both equal (n).
Correct answer: D. (^{n}C_2=\frac{n!}{2!(n-2)!})
Explanation: In the combination formula after cancelling ((n-2)!) the term (n(n-1)) remains. In exams remember (^{n}C_2) as the pair-making formula.
Correct answer: C. Filling two ordered positions
Explanation: There are (n) choices for the first place and (n-1) for the second. In exams use the multiplication principle for ordered pairs.
Correct answer: B. (^{9}C_2)
Explanation: By (^{n}C_r=^{n}C_{n-r}) we get (^{9}C_7=^{9}C_2). In exams use the smaller index to reduce calculation.
Correct answer: A. Not choosing the special object
Explanation: Choosing (r) from remaining (n-1) after excluding one fixed object is (^{n-1}C_r). In exams split cases into included and excluded.
