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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. Both rotation and reflection are considered the same
Explanation: After removing circular duplicates, mirror images are also the same. In exams divide by (2) for reflection in bracelet problems.
Correct answer: A. Rotations are considered the same in circular seating
Explanation: In a circle, fixing one person removes rotational overcount. In exams use the one-fixed method for round tables.
Correct answer: A. (2!(n-2)!)
Explanation: Treating (A) and (B) as one block gives (n-1) circular objects. In exams remember ((n-2)!) in circular block counts.
Correct answer: B. (^{n}P_r)
Explanation: Positions are ordered and objects are not repeated, so falling choices arise. In exams think of permutations when positions are distinct.
Correct answer: A. Each position has (6) independent choices
Explanation: Choices do not decrease when repetition is allowed. In exams use the power formula when repetition is allowed.
Correct answer: A. Because if the first digit is (0), the number will not remain (4)-digit
Explanation: A leading zero does not form a valid (4)-digit number. In exams check the first-place condition first in digit arrangements.
Correct answer: A. When repetition is not allowed and order is important
Explanation: Without repetition, choices decrease, so (^{n}P_r) is needed. In exams do not apply powers without reading the repetition condition.
Correct answer: B. (10^4)
Explanation: Each distinct prize has (10) independent student choices. In exams use the power rule for distinct prizes with repetition allowed.
Correct answer: C. (^{10}P_4)
Explanation: Prizes are different and recipients cannot repeat, so it is an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Correct answer: A. ({}^{13}C_9)
Explanation: Identical prize distribution is stars and bars, with (4) stars and (9) bars. In exams connect identical prizes with combinations with repetition.
Correct answer: C. (^{8}C_5)
Explanation: Prizes are identical and only (5) students need to be chosen. In exams treat at most one with identical prizes as simple selection.
Correct answer: A. Because the consecutive ratio is first greater than (1) and later less than (1)
Explanation: The ratio (\frac{n-r}{r+1}) shows the transition. In exams check the trend of binomial coefficients by ratios.
Correct answer: A. ({}^{n}C_{\frac{n}{2}})
Explanation: For even (n), binomial coefficients peak at the middle. In exams identify the middle term using symmetry and ratio.
Correct answer: B. ({}^{n}C_{\frac{n-1}{2}}) and ({}^{n}C_{\frac{n+1}{2}})
Explanation: For odd (n), the two middle indices are complementary and equal. In exams remember two largest terms for the odd case.
Correct answer: A. (r+s=n)
Explanation: In combination symmetry, unequal equal-values come from complementary indices. In exams check sum (n) for equal combinations.
Correct answer: A. Because (^{n}P_r) usually changes by extra positive factors as (r) increases
Explanation: Permutations do not have complement symmetry, and changing length changes the count. In exams do not apply combination symmetry to permutations.
Correct answer: A. Combination is obtained by dividing permutation count by (r!)
Explanation: The numerator is (^{n}P_r) and the denominator removes order. In exams use this form for quick numerical cancellation.
Correct answer: A. Because (r!\geq1)
Explanation: Permutation counts all orders of each selected group. In exams equality is also possible for (r=0) or (r=1).
Correct answer: A. ({}^{n+1}C_{r+1})
Explanation: Adjacent lower indices (r) and (r+1) give the (r+1) term of the next row. In exams identify Pascal pattern quickly.
Correct answer: A. ({}^{12}C_4+{}^{12}C_5={}^{13}C_5)
Explanation: By Pascal identity, the same upper index with adjacent lower indices combines. In exams first combine the adjacent pair.
Correct answer: A. By adding ((1+1)^n) and ((1-1)^n)
Explanation: To separate even-index coefficients, two binomial substitutions are added. In exams use (x=1) and (x=-1) for even-odd sums.
Correct answer: B. (n-r+1)
Explanation: Filling one more position gives (n-r+1) choices. In exams identify the new last factor in the next permutation term.
Correct answer: A. Counting by marking one selected member
Explanation: Choosing the marked member first gives (n) choices and each group is counted (r) times. In exams treat the denominator (r) as overcount.
Correct answer: C. (\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r})
Explanation: Canceling factorial forms gives (\frac{n}{r}). In exams use the ratio method for adjacent upper and lower indices.
Correct answer: A. (^{n+1}C_r)
Explanation: Adjacent terms of the same row give the middle term in the next row. In exams apply Pascal identity when (r) and (r-1) appear together.
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