Derivations of formulas and their connections

Correct answer: B. (^{4}P_3)

Explanation: Without repetition the choices are (4), (3), (2) so (^{4}P_3) is correct. In exams use permutation when there is no repetition.

Correct answer: A. (^{n}P_r) has no repetition and (n^r) allows repetition

Explanation: In permutation a chosen object does not appear again but in (n^r) every position has all choices. In exams always read the repetition condition.

Correct answer: A. Because it counts only selections

Explanation: A combination counts groups or selections, not ordered placements. If the selected objects are A and B, the selections AB and BA represent the same two-element set because the membership of the group has not changed. Permutations, in contrast, treat AB and BA as different arrangements. This is why nCᵣ is used when order is irrelevant, whereas nPᵣ is used when order matters; in fact, nPᵣ = nCᵣ × r!. Option A gives the correct meaning. Option B describes permutations, option C incorrectly restricts combinations to circles, and option D confuses combinations with unrestricted repeated choices.

Correct answer: A. Person (A) to (B) and (B) to (A) is the same handshake

Explanation: A handshake is an unordered pair so (AB) and (BA) are not counted separately. In exams use combination when a pair has no direction.

Correct answer: B. 6

Explanation: Use the complementary identity \(^{n}C_r=^{n}C_{n-r}\). Thus, \(^{10}C_4=^{10}C_{10-4}=^{10}C_6\), so \(m=6\). Although \(m=4\) also gives the same value, it is excluded in the question. Exam tip: for a fixed \(n\), \(r\) and \(n-r\) always give equal combinations.

Correct answer: A. Order inside chosen and not chosen groups is irrelevant

Explanation: In (n!) internal orders inside chosen and unchosen groups are counted extra. In exams connect group division with factorial denominators.

Correct answer: A. Counting in two ways by marking one chosen object

Explanation: A marked selection can start by choosing one of (n) objects and then choosing the remaining (r-1). In exams identify double counting identities.

Correct answer: B. (\frac{n-r+1}{r})

Explanation: Dividing the factorial forms leaves (\frac{n-r+1}{r}). In exams use the ratio formula for adjacent terms.

Correct answer: B. (n-r+1)

Explanation: When the (r)th position is added the new factor is (n-r+1). In exams the ratio of consecutive permutations gives the last factor.

Correct answer: A. First choose then arrange the chosen objects

Explanation: Permutation includes both selection and arrangement. In exams multiply by (r!) to connect (P) with (C).

Correct answer: B. (^{8}C_3\times3!)

Explanation: First choose (3) books and then order them in (3!) ways. In exams add arrangement after selection when ordered placement appears.

Correct answer: A. (^{7}P_2)

Explanation: Captain and vice captain are different posts so order is important. In exams use permutation for different posts.

Correct answer: C. (^{7}C_2)

Explanation: There is no post or order in choosing members so combination is used. In exams do not count order when the phrase only select appears.

Correct answer: B. (n!)

Explanation: The choices are (n) for the first position then (n-1) down to (1). In exams connect complete arrangement with factorial.

Correct answer: B. Multiplication principle

Explanation: Choices for each position decrease after previous selection and are multiplied together. In exams use the multiplication principle in arrangement.

Correct answer: A. Number of unordered pairs of vertices among (n) vertices

Explanation: An edge is formed by an unordered pair of two vertices. In exams use (^{n}C_2) for pairs without direction.

Correct answer: A. Ordered pair of (2) from (n) objects

Explanation: In a directed pair (AB) and (BA) are different so permutation is used. In exams choose (P) when direction or order appears.

Correct answer: B. (2!)

Explanation: The (2!) orders of two selected objects are counted extra in permutation. In exams the divide factor for (r=2) is (2!).

Correct answer: A. (4!)

Explanation: The (4!) internal orders of four selected objects must be removed. In exams apply (^{n}C_r=\frac{^{n}P_r}{r!}).

Correct answer: B. (4!)

Explanation: The four chosen objects can be ordered in (4!) ways. In exams multiply by (r!) when going from (C) to (P).

Correct answer: A. The unchosen group is fixed by the chosen group

Explanation: An (r)-selection fixes an ((n-r))-rejection with it. In exams view choose and reject as a pair.

Correct answer: A. (^{11}C_8=^{11}C_3)

Explanation: Choosing (8) means leaving (3) so both combinations are equal. In exams replace large (r) with the smaller complement.

Correct answer: A. \(^{n}C_r = {}^{n}C_{n-r}\)

Explanation: Choosing \(r\) objects from \(n\) uniquely determines the \(n-r\) objects left out, so \(^{n}C_r=^{n}C_{n-r}\). For example, \(^{5}C_2=10=^{5}C_3\). Exam tip: replace \(r\) by \(n-r\) to use the smaller value in a combination.

Correct answer: A. \(^{n}P_r=r!\,{}^{n}C_r\)

Explanation: In a permutation, the order of the selected \(r\) objects matters. Each combination can be arranged in \(r!\) different orders, so \(^{n}P_r=r!\,{}^{n}C_r\). Option B incorrectly divides by \(r!\). Exam tip: when order matters, multiply combinations by \(r!\).

Correct answer: A. Once the group of (r) is chosen the other group is automatically fixed

Explanation: Choosing (r) objects fixes the remaining (n-r) objects. In exams connect group division with combination.