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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. (^{n}C_m2^{n-m})
Explanation: First place the (m) special members in the second part and give free choices to the remaining (n-m) members. In exams choose the fixed marked set first in such sums.
Correct answer: B. (^{n}C_3 2^{n-3})
Explanation: Choose the (3) marked members first and let each of the remaining (n-3) members enter the subset or not. In exams count the inner selection first.
Correct answer: A. (nx(1+x)^{n-1})
Explanation: Differentiate ((1+x)^n) and multiply by (x) to obtain this sum. In exams think of the derivative method when a factor (r) appears.
Correct answer: A. (n(n-1)x^2(1+x)^{n-2})
Explanation: Two differentiations produce the factor (r(r-1)), and (x^2) restores the power. In exams use the second derivative for (r(r-1)).
Correct answer: A. (r^3=r(r-1)(r-2)+3r(r-1)+r)
Explanation: Breaking powers into falling factorials allows standard binomial sums. In exams write (r^3) in falling form instead of expanding directly.
Correct answer: A. Choosing the large group first or choosing the marked (s)-group first is the same task
Explanation: Both sides count the same selection with (s) special members inside an (r)-group. In exams change the order of nested selection.
Correct answer: A. (\frac{n!}{a!b!c!(n-a-b-c)!})
Explanation: In sequential selection the internal order of each selected block is removed. In exams form a multinomial denominator when many labelled groups appear.
Correct answer: A. (2!\cdot2!)
Explanation: Interchanging the two size (2) groups and the two size (4) groups gives duplicates. In exams divide extra by factorials of equal-size unlabelled groups.
Correct answer: B. (\frac{15!}{(5!)^3 3!})
Explanation: All three groups have the same size and are unlabelled, so group order (3!) is also removed. In exams remember the extra (3!) for equal unlabelled groups.
Correct answer: A. (^{r}C_k\sum_{i=0}^{r-k}(-1)^i{}^{r-k}C_i(r-k-i)^n)
Explanation: First choose the empty boxes, then distribute onto the remaining boxes. In exams use selection plus inclusion-exclusion for exactly empty boxes.
Correct answer: A. (4^n-4\cdot3^n+6\cdot2^n-4)
Explanation: Empty boxes are subtracted and added by inclusion-exclusion. In exams interpret onto as every labelled box being non-empty.
Correct answer: A. \(\frac{1}{3!}\left(3^n-3\cdot2^n+3\right)\)
Explanation: First count (3) labelled non-empty groups, then remove the (3!) label permutations. In exams divide the labelled count for unlabelled groups.
Correct answer: A. (^{19}C_3)
Explanation: After removing the minimum sum (14), (16) remains, so ({}^{16+4-1}C_{3}) is obtained. In exams subtract unequal lower bounds first.
Correct answer: B. (1)
Explanation: Since the total is (24) and the maximum of each of the three variables is (8), only ((8,8,8)) is possible. In exams check extreme feasibility before inclusion-exclusion.
Correct answer: A. (^{20}C_3-4{}^{14}C_3+6{}^{8}C_3-4{}^{2}C_3)
Explanation: A violation starts at (x_i\geq6), so inclusion-exclusion applies. In exams use shift (6) for upper bound (5).
Correct answer: A. (^{5}C_3{}^{11}C_2)
Explanation: Choose the (3) positive variables first, then split (12) into (3) positive parts. In exams use choose variables plus positive stars and bars for exactly positive variables.
Correct answer: A. (^{5}C_2{}^{19}C_2)
Explanation: First choose the empty boxes, then distribute positively into the remaining (3) boxes. In exams split exactly empty into box selection and positive distribution.
Correct answer: A. Derangement inclusion-exclusion formula
Explanation: Comparing the alternating factorial expression for derangements gives this recurrence. In exams remember the inclusion-exclusion form for derangement recurrence.
Correct answer: A. (^{7}C_2D_5)
Explanation: First choose the (2) objects that stay fixed and derange the remaining (5). In exams use choose fixed plus derange rest for exactly fixed points.
Correct answer: A. (D_6=265)
Explanation: This is a derangement of (6) objects and (D_6=265). In exams treat letter-envelope mismatch as derangement.
Correct answer: A. (2(n-k-1)(n-2)!)
Explanation: The positions of (A,B) are at distance (k+1), and there are (2) choices for order. In exams count positions first in fixed-gap problems.
Correct answer: A. (2\cdot6\cdot8!)
Explanation: There are (10-3-1=6) position choices and (2) choices for the order of (A,B). In exams count position pairs for between conditions.
Correct answer: A. (\frac{9!}{3!})
Explanation: Only (1) of the (3!) relative orders of those (3) books is allowed. In exams divide total by (k!) for fixed relative order.
Correct answer: A. (\frac{12!}{8})
Explanation: Three independent before-after restrictions divide the count by (2^3). In exams divide by a power of (2) for independent pairs.
Correct answer: A. Fix (A) and check the two possible circular positions of (B)
Explanation: After removing circular rotation, positions with fixed distance are counted. In exams fix one object first for circular distance.
