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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. Removing orders inside two groups
Explanation: In (n!), internal orders of chosen and unchosen groups are counted separately. In exams remove internal order in group division.
Correct answer: B. (^{n}C_r)
Explanation: Once one group is chosen, the other group is fixed, so (^{n}C_r) is enough. In exams count only one selection when the complement is fixed.
Correct answer: A. (^{8}C_3\times3)
Explanation: First there are (^{8}C_3) ways to form the team and (3) choices for captain. In exams multiply when an extra role follows selection.
Correct answer: C. (^{8}P_3\div2!)
Explanation: One of the three is distinguished as captain, so removing the order of the remaining (2) members gives (^{8}P_3\div2!). In exams reduce overcounting by checking the roles.
Correct answer: A. (2!2!3!)
Explanation: Internal orders of each repeated group are not different, so divide by (2!2!3!). In exams treat repeated letters as separate groups.
Correct answer: A. Division principle due to repeated letters
Explanation: Interchanging identical letters does not make a new arrangement. In exams write frequencies of repeated letters and divide by factorials.
Correct answer: A. (\frac{1}{8})
Explanation: The circular count is (7!) and the linear count is (8!), so the ratio is (\frac{7!}{8!}=\frac{1}{8}). In exams rotations are removed in circular counting.
Correct answer: B. (\frac{5!}{2})
Explanation: Removing rotation first gives (5!), then reflection being same makes us divide by (2). In exams check both rotation and reflection in necklaces.
Correct answer: C. (^{5}C_2\times{}^{4}C_2)
Explanation: The selections of boys and girls are independent and order does not matter in a committee. In exams multiply combinations for category-wise selection.
Correct answer: A. (^{4}C_2{}^{5}C_2+^{4}C_3{}^{5}C_1+^{4}C_4{}^{5}C_0)
Explanation: The number of girls can be (2), (3), or (4), and the cases are added. In exams write valid cases separately in at least questions.
Correct answer: A. (2^{n-1})
Explanation: Even and odd indexed combination sums are equal and the total sum is (2^n). In exams remember ((1-1)^n) for the even-sum identity.
Correct answer: B. (2^{n-1})
Explanation: The sum of odd indexed combinations equals the even indexed sum and is (2^{n-1}). In exams prove it using the alternating identity.
Correct answer: C. (6)
Explanation: Different lower indices must be complementary, so (2+4=n). In exams quickly check complementary indices in equal (C) terms.
Correct answer: A. Because (2!=2)
Explanation: Two objects have (2!) orders and (2!=2). In exams remember the relation between ordered and unordered counts in pairs.
Correct answer: A. It is (^{7}C_4\times4!)
Explanation: First choose (4) letters and then arrange them in (4!) orders. In exams order is important in words.
Correct answer: A. (4\times5\times5)
Explanation: The first place cannot have (0), so it has (4) choices and the remaining places have (5) choices. In exams check the leading zero condition first.
Correct answer: A. (5\times{}^{5}P_3)
Explanation: There are (5) non-zero choices for the first digit and then the remaining (3) places are filled from (5) remaining digits. In exams treat the first place as a separate case.
Correct answer: A. (n=2r-1)
Explanation: Different lower indices are complementary, so (r+(r-1)=n). In exams this relation is useful for adjacent equal combinations.
Correct answer: A. (n-r+1=r)
Explanation: The ratio is (\frac{n-r+1}{r}), and setting it equal to (1) gives (n-r+1=r). In exams form an equation using the ratio identity.
Correct answer: A. (n-r+1)
Explanation: In the recurrence, the extra factor is the choices for the (r)th place, which is (n-r+1). In exams treat the multiplier as next-place choices.
Correct answer: A. (r!(n-r)!)
Explanation: (^{n}C_{n-r}=\frac{n!}{(n-r)!r!}), which gives the same denominator. In exams verify symmetry using factorials.
Correct answer: A. The same two factorials remain in the denominator
Explanation: In (\frac{n!}{r!(n-r)!}), changing the order of (r!) and ((n-r)!) does not change the value. In exams order does not matter in a product denominator.
Correct answer: A. (^{11}C_4)
Explanation: Once (4) selected objects are chosen, the (7) rejected objects are fixed. In exams connect selected-rejected split with combination.
Correct answer: A. Unchosen objects
Explanation: In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.
Correct answer: A. Because (1!=1)
Explanation: One selected object has only (1) order. In exams permutation and combination both equal (n) when (r=1).
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