Derivations of formulas and their connections

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

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.

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.

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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

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

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

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

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

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

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.

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 \(r!\).

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