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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. (n=2r-3)
Explanation: Different lower indices are complementary, so (r+(r-3)=n). In exams solve equal combinations using the complement rule.
Correct answer: B. (n-r=3(r+1))
Explanation: The ratio is (\frac{n-r}{r+1}), and it is set equal to (3). In exams form the equation directly from the ratio formula.
Correct answer: A. \(n-r+1=4r\)
Explanation: For adjacent combinations, \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Hence, \(\frac{n-r+1}{r}=4\) gives \(n-r+1=4r\). Therefore, option A is correct. Option D incorrectly omits the \(+1\), which arises from the \(^{n}C_{r-1}\) term. Exam tip: in ratios of adjacent combinations, carefully check that the numerator is \(n-r+1\).
Correct answer: C. (^{12}C_5)
Explanation: A pentagon needs an unordered selection of (5) points. In exams do not count the order of points when forming a shape.
Correct answer: D. (^{12}P_2)
Explanation: In a directed segment, changing start and end changes the object. In exams use permutation when direction exists.
Correct answer: C. (^{12}C_2)
Explanation: In an ordinary segment, the order of endpoints does not matter. In exams use (^{n}C_2) for unordered pairs.
Correct answer: A. Circular count (=\frac{9!}{9})
Explanation: Each circular arrangement is counted (9) times in the linear count due to rotations. In exams treat rotations as extra count in a circle.
Correct answer: B. (\frac{7!}{2})
Explanation: First removing rotations gives (7!), then mirror images being the same makes us divide by (2). In exams always check reflection in necklace problems.
Correct answer: A. (4!2!)
Explanation: Internal interchanges of repeated letters do not create new arrangements. In exams divide by the factorial of each repeated group.
Correct answer: B. (\frac{10!}{4!3!})
Explanation: Internal orders of two identical groups are not different, so divide by (4!3!). In exams multiply the factorials in the denominator.
Correct answer: B. This is impossible because (r!>1)
Explanation: (^{n}P_r=^{n}C_r r!), and for (r>1), (r!>1). In exams verify equality of (P) and (C) using factorials.
Correct answer: B. 7
Explanation: The relation between permutations and combinations is \(^{n}P_r=\,^{n}C_r\times r!\). Hence, \(^{n}C_r=\frac{^{n}P_r}{r!}\). Here the denominator is 5040, so \(r!=5040\). Since \(7!=5040\), \(r=7\). For instance, \(6!=720\), so 6 is not correct. Exam tip: in such questions, equate the given denominator directly to \(r!\) and identify the factorial.
Correct answer: B. (5)
Explanation: The direct answer is option B, \(r=5\). Use the identity \(^{n}P_r=^{n}C_r\,r!\). The question gives \(^{n}P_r=120\times{}^{n}C_r\). After comparing the common factor \(^{n}C_r\), we get \(r!=120\). Since \(5!=5\cdot4\cdot3\cdot2\cdot1=120\), the value is \(r=5\). Option A, 4, would give \(4!=24\), not 120. Option B, 5, gives exactly 120 and is correct. Option C, 6, gives \(6!=720\). Option D, 7, gives \(7!=5040\). The identity comes from first choosing r objects, counted by the combination, and then arranging those chosen objects in \(r!\) orders. Exam cue: the numerical multiplier between \(^{n}P_r\) and \(^{n}C_r\) is always \(r!\), provided the common combination factor is nonzero.
Correct answer: B. (7)
Explanation: Complementary indices have sum (20), so (a+a+6=20) and (a=7). In exams form a linear equation from equal combinations.
Correct answer: C. (7)
Explanation: Different lower indices are complementary, so (x+2x=21) and (x=7). In exams solve equal (C) terms using the complement rule.
Correct answer: B. (7)
Explanation: This is the sum of non-empty selections (2^n-1), and (2^7-1=127). In exams subtract (1) when the empty selection is removed.
Correct answer: C. (9)
Explanation: The total sum is (2^n), and (2^9=512). In exams connect the sum of all combinations with a power of (2).
Correct answer: B. (9)
Explanation: The even indexed sum is (2^{n-1}), and (2^8=256) gives (n=9). In exams remember that even and odd sums are equal.
Correct answer: A. To remove internal orders of chosen and unchosen groups
Explanation: In the (n!) count, the orders inside both groups are counted extra. In exams understand both corrections (r!) and ((n-r)!).
Correct answer: A. (\frac{n!}{a!b!c!})
Explanation: Order inside each labelled group is irrelevant, so divide by (a!b!c!). In exams do not divide extra among labelled groups themselves.
Correct answer: A. (\frac{^{10}C_5}{2!})
Explanation: After choosing the first (5)-group the second is fixed, and interchanging the two groups creates duplicate count. In exams divide by (2!) for equal groups.
Correct answer: A. False, because the correct multiplier is (4!=24)
Explanation: (^{n}P_r=^{n}C_r r!), so for (r=4) the multiplier is (24). In exams verify the multiplier using factorials.
Correct answer: A. (^{n}P_r) counts selection along with arrangement of selected objects
Explanation: Permutation is (r!) times combination because the selected objects must be ordered. In exams remember (P=C\times r!) conceptually.
Correct answer: B. (n-3=2\times4)
Explanation: The ratio is (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}), so for (r=4), (\frac{n-3}{4}=2). In exams use the ratio formula for adjacent combinations.
Correct answer: C. (4!)
Explanation: The relation is (^{n}P_r=^{n}C_r\times r!), so here the factor is (4!). In exams add the arrangements of selected objects when going from (C) to (P).
