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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. Fixing the largest chosen element
Explanation: This is the hockey-stick identity and the largest selected element creates cases. In exams use the last-element method for such staircase sums.
Correct answer: B. Treat it as a shifted form of the hockey-stick identity
Explanation: The upper and lower indices increase together so it is a diagonal sum. In exams think of hockey-stick when a diagonal combination sum appears.
Correct answer: B. Choosing a subset and selecting an unordered marked pair inside it
Explanation: Choose the marked pair first and choose the remaining elements freely. In exams think of pair marking when ({}^{r}C_2) appears.
Correct answer: C. (^{n}C_3 2^{n-3})
Explanation: Choose the (3) marked members first and freely choose the remaining (n-3) members. In exams count the inner combination first.
Correct answer: D. (2{}^{r}C_2+3{}^{r}C_3)
Explanation: If a pair is marked and then one member is marked, cases occur inside or outside the pair. In exams split products of marks into cases.
Correct answer: A. (^{n}C_c{}^{n-c}C_{b-c}{}^{n-b}C_{a-b})
Explanation: Choose the innermost (c) elements first and then add layers. In exams count nested choices from inside to outside.
Correct answer: A. (s\leq r\leq n)
Explanation: The smaller selected set must lie inside the larger selected set. In exams first check valid index order in nested combinations.
Correct answer: B. (0)
Explanation: Differentiate ((1+x)^n) and put (x=-1) to get zero. In exams use derivative plus (x=-1) for alternating weighted sums.
Correct answer: C. (0)
Explanation: After the second derivative and putting (x=-1), ((1-1)^{n-2}) appears. In exams check zero when the falling factor is below the degree.
Correct answer: A. (r^2=r(r-1)+r)
Explanation: This split connects first and second weighted sums. In exams convert powers into falling factorials.
Correct answer: C. (n(n+1)2^{n-2})
Explanation: Using (r^2=r(r-1)+r) adds two standard sums. In exams simplify the final form to (n(n+1)2^{n-2}).
Correct answer: A. (n^2(n+3)2^{n-3})
Explanation: Using (r^3=r(r-1)(r-2)+3r(r-1)+r) gives this form. In exams use falling factorial decomposition for cubic sums.
Correct answer: C. (2!)
Explanation: Two groups have the same size (a), so interchanging them gives duplicates. In exams divide extra by factorial for equal-size unlabelled groups.
Correct answer: A. (3!3!4!4!2!2!)
Explanation: Both internal orders and equal-size group swaps are removed. In exams write both group sizes and equal-group factorials in the denominator.
Correct answer: A. (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right)) / (^{r}C_1\left((r-1)^n-{}^{r-1}C_1(r-2)^n+\cdots\right)
Explanation: Choose the empty box first and then distribute onto the remaining (r-1) boxes. In exams use choose empty plus onto count for exactly non-empty boxes.
Correct answer: A. (^{4}C_2(2^n-2))
Explanation: Choose the empty boxes first and then the remaining two boxes must both be non-empty. In exams apply onto to the remaining boxes in exactly-empty cases.
Correct answer: A. (3^n-3\cdot2^n+3)
Explanation: Empty-box cases are removed by inclusion-exclusion. In exams count non-empty labelled boxes like onto functions.
Correct answer: B. (^{15}C_3)
Explanation: After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.
Correct answer: A. (^{21}C_2-3{}^{13}C_2+3{}^{5}C_2)
Explanation: The upper violation is (x_i\geq8) and inclusion-exclusion is applied. In exams use shift (8) for upper bound (7).
Correct answer: A. (^{4}C_1{}^{15}C_2)
Explanation: Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.
Correct answer: A. (^{6}C_3{}^{17}C_2)
Explanation: First choose the (3) non-empty boxes and then split (18) balls into (3) positive parts. In exams use choose boxes plus positive stars and bars for exactly non-empty.
Correct answer: B. When the first object and its target object swap places
Explanation: If two objects swap with each other, the remaining (n-2) objects are deranged. In exams separate swap and non-swap cases in derangement recurrence.
Correct answer: A. (1854)
Explanation: Using the derangement recurrence, (D_7=6(D_6+D_5)=1854). In exams remember small (D_n) values through recurrence.
Correct answer: A. (^{8}C_3D_5)
Explanation: First choose the (3) fixed objects and derange the remaining (5) objects. In exams use choose fixed plus derange rest for exactly fixed points.
Correct answer: A. (^{7}C_2D_5)
Explanation: Choose the correct letters and derange the remaining letters into wrong envelopes. In exams solve exactly correct letters using the fixed-point formula.
