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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: B. (^{7}C_5\cdot5!)
Explanation: After (6) boys, (7) gaps are formed, and (5) girls are arranged in them. In exams use the gap method for no two together.
Correct answer: B. (2\cdot5!\cdot5!)
Explanation: There are (2) patterns for the starting gender and both groups arrange in (5!) ways. In exams do not forget the (2) patterns for equal alternate groups.
Correct answer: B. (\frac{7!}{2})
Explanation: Removing circular rotations gives (7!), and divide by (2) when reflection is the same. In exams read the mirror-image condition in necklace problems.
Correct answer: A. One person is fixed to remove rotations
Explanation: Rotations give the same arrangements at a round table. In exams write ((n-1)!) for circular seating.
Correct answer: B. (4!\cdot2^5)
Explanation: The circular arrangement of (5) couple-blocks is (4!) and each block has (2) internal orders. In exams use one less factorial for circular blocks.
Correct answer: A. Because (A) and (R) appear twice each
Explanation: Interchanging identical letters does not create a new arrangement. In exams put factorials of repeated letters in the denominator.
Correct answer: A. (3!3!2!)
Explanation: (S) appears three times, (T) appears three times, and (I) appears twice. In exams count letter frequencies before word arrangement.
Correct answer: A. (\frac{8!}{3!})
Explanation: Only (1) of the (3!) possible relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.
Correct answer: B. (\frac{10!}{3!})
Explanation: Only one of the (3!) relative orders of these (3) people is valid. In exams handle fixed order restriction by symmetry division.
Correct answer: B. (1,3,5)
Explanation: For an odd number, the last digit must be odd. In exams decide the unit-place condition first in digit problems.
Correct answer: B. (5\cdot6\cdot6)
Explanation: The first digit cannot be (0), and the remaining two places have (6) choices. In exams handle the leading-zero restriction separately.
Correct answer: A. Each position has (7) independent choices
Explanation: When repetition is allowed, a selected symbol remains available again. In exams use the power rule for independent positions.
Correct answer: C. (^{7}P_5)
Explanation: Order matters in passwords and repetition is not allowed. In exams use permutation for ordered slots without repetition.
Correct answer: A. Because (b) is chosen from (r) brackets
Explanation: To form (b^r), choose (r) brackets from (n) brackets. In exams connect binomial coefficients with bracket selection.
Correct answer: A. Adding all possible subset sizes
Explanation: The left side adds subsets of every size and the right side gives choose-or-not choices for each object. In exams count subset identities in two ways.
Correct answer: A. Choosing r objects from n objects can be counted by choosing the \(n-r\) objects left unchosen.
Explanation: Each r-element selection has exactly one complementary group of \(n-r\) unselected elements. This one-to-one correspondence gives equal counts. Option B describes permutations, where order matters. Exam tip: identify the selected and unselected complementary sets.
Correct answer: A. Marking one selected member
Explanation: Choose a subset and mark one member, or first choose the marked member and freely choose the rest. In exams treat the factor (r) as a marked choice.
Correct answer: A. (^{n}C_2 2^{n-2})
Explanation: First choose the marked pair, then freely choose from the remaining (n-2) objects. In exams choose marked objects first in nested combination sums.
Correct answer: B. Vandermonde identity
Explanation: When choosing (r) objects from two groups, (k) objects are taken from the first group. In exams identify Vandermonde in two-group selection.
Correct answer: A. (\frac{n!}{a!b!(n-a-b)!})
Explanation: Choosing (a) objects first and then (b) from the remaining objects is like labelled grouping. In exams convert sequential selections into factorial division.
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