Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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. Choosing total (r) objects from two separate groups
Explanation: Choose (k) from the first group and (r-k) from the second, then add all cases. In exams identify Vandermonde in two-group selection.
Correct answer: B. Choosing total (n) objects from two groups of size (n)
Explanation: Choosing (n-r) from the second group equals ({}^{n}C_{n-r}={}^{n}C_r). In exams connect square sums with two equal groups.
Correct answer: B. (n!)
Explanation: Choose (r) objects first, then arrange both chosen and unchosen lists. In exams factorial cancellation also gives this quickly.
Correct answer: A. Marking one member in a chosen (r)-group
Explanation: The left side chooses a group and marks one member, while the right side chooses the marked member first. In exams treat the factor (r) as a marked choice.
Correct answer: A. Ordered two marked members in the selected group
Explanation: In an (r)-group, the first and second marked members are chosen with order. In exams think of ordered marking when you see (r(r-1)).
Correct answer: A. After choosing the marked member, the remaining (n-1) members are freely chosen or left
Explanation: There are (n) choices for the marked member and two choices for each remaining member. In exams connect the extra (r) in binomial sums with marking.
Correct answer: B. (n(n-1)2^{n-2})
Explanation: There are (n(n-1)) ways to choose two ordered marked members and the remaining (n-2) are chosen freely. In exams two marks lead to (2^{n-2}).
Correct answer: C. (n(n+1)2^{n-2})
Explanation: Write (r^2=r(r-1)+r) and add two standard sums. In exams splitting (r^2) is the fastest method.
Correct answer: A. (^{n}C_k2^{n-k})
Explanation: Choose the (k) marked members first, then each of the remaining (n-k) members may or may not enter the subset. In exams choose the marked set first in nested selection.
Correct answer: A. (^{n}C_t{}^{n-t}C_{s-t}{}^{n-s}C_{r-s})
Explanation: First choose the innermost (t) members, then add (s-t), then (r-s) members. In exams count nested choices backward by layers.
Correct answer: B. (\frac{n!}{a!b!c!})
Explanation: Labelled groups are fixed and order inside each group is not counted. In exams put group-size factorials in the multinomial denominator.
Correct answer: B. (\frac{12!}{4!4!4!3!})
Explanation: All three groups have equal size and are unlabelled, so the (3!) interchange of groups is also removed. In exams divide extra by factorial for equal unlabelled groups.
Correct answer: A. (a+b+c=n)
Explanation: Choose (a) first, then (b) from the remaining, and (c) is automatically fixed. In exams match sequential selection with factorial form.
Correct answer: A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n)
Explanation: Cases with empty boxes are removed from total functions by inclusion-exclusion. In exams read onto as every box being non-empty.
Correct answer: A. (3^n-3\cdot2^n+3)
Explanation: Three empty-box choices are subtracted, then over-subtraction of two empty boxes is added. In exams apply inclusion-exclusion carefully for onto distribution.
Correct answer: A. (2^n-2)
Explanation: From total (2^n) assignments, the two all-in-one empty-box cases are invalid. In exams subtract empty cases separately for labelled boxes.
Correct answer: A. (\frac{2^n-2}{2})
Explanation: In the labelled count, interchanging the two groups counts each division twice. In exams divide by (2) for two unlabelled groups.
Correct answer: A. Stars and bars
Explanation: This is multiset selection and the answer is ({}^{n+r-1}C_r). In exams handle repetition with no order by the bars method.
Correct answer: B. (^{20}C_3)
Explanation: Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.
Correct answer: A. C(10,2)
Explanation: The governing concept is the stars-and-bars method with lower bounds. First remove the compulsory minimum values by setting x₁=3+y₁, x₂=4+y₂, and x₃=5+y₃, where y₁,y₂,y₃ are non-negative integers. Substitution gives y₁+y₂+y₃=20−3−4−5=8. The number of non-negative integer solutions of y₁+y₂+y₃=8 is C(8+3−1,3−1)=C(10,2). Therefore option A is correct. Option B incorrectly uses 8 as the top number without adding the two separators. Options C and D incorrectly add the lower bounds instead of subtracting the compulsory amounts. The count assumes integer solutions, as stated.
Correct answer: A. (^{4}C_2{}^{17}C_1)
Explanation: Choose the zero variables first, then the remaining two variables form a positive sum of (18). In exams use positive distribution for exactly-zero cases.
Correct answer: A. (^{17}C_2-3{}^{10}C_2+3{}^{3}C_2)
Explanation: Cases with (x_i\geq7) are subtracted and double violations are added. In exams shift by (7) for bounded solutions.
Correct answer: A. (^{13}C_3-4{}^{8}C_3+6{}^{3}C_3)
Explanation: A violation begins with (x_i\geq5), so inclusion-exclusion applies. In exams use a subtract shift of (5) for upper limit (4).
Correct answer: A. To choose the wrong position of the first letter
Explanation: The first letter can go to (n-1) positions other than its original position. In exams watch the first object's wrong choice in derangement recurrence.
Correct answer: A. Inclusion-exclusion
Explanation: Permutations with fixed points are subtracted and added alternately. In exams think of inclusion-exclusion when no object is in its correct place.
