Derivations of formulas and their connections

For n distinct objects in a line, there are n! linear arrangements. When the objects are placed around a circle, rotating the entire circle does not create a genuinely new arrangement; it only changes the position chosen as the starting point. Each circular arrangement is therefore represented by exactly n linear arrangements, one for each possible rotation. Dividing n! by n removes this repeated counting and gives n!/n = (n-1)! circular arrangements. Option A expresses this principle. The objects do not become identical, order still matters around the circle, and the formula is not based on selecting only two objects.

The connection between permutations and combinations explains the division by 2!. The expression ⁿP₂ counts ordered selections: for two distinct objects A and B, it counts AB and BA separately. In an unordered pair, however, AB and BA represent the same pair because no direction or first position is specified. Every pair is therefore counted exactly 2! = 2 times in ⁿP₂. Dividing by 2! removes this duplicate ordering and gives ⁿC₂. Option B states this accurately. The objects need not be identical, n choices are not the reason, and circularity is irrelevant to an ordinary pair.

For a set containing n elements, the term ⁿCₖ counts the subsets containing exactly k elements. Summing from k = 0 through k = n counts every subset once, grouped according to its size; the k = 0 term is the empty subset and the k = n term is the full set. Independently, each element has exactly two choices in forming a subset: include it or exclude it. By the multiplication principle, n independent two-way choices produce 2 × 2 × ... × 2 = 2ⁿ subsets. Therefore option B is correct. n! and ⁿPₙ count arrangements, while n² does not represent the binary choice for each element.

Use the factorial definition of permutations for the next lower index: ⁿPᵣ₊₁ = n!/[n-(r+1)]! = n!/(n-r-1)!. The given expression is ⁿPᵣ = n!/(n-r)!. Since (n-r)! = (n-r)(n-r-1)!, the next permutation value equals [n!/(n-r)!](n-r) = ⁿPᵣ(n-r). Thus option A is correct. In counting terms, after r positions have been filled, n-r objects remain available for the next position, which explains the multiplying factor. Option B incorrectly uses r+1, option C reverses the relationship by dividing, and option D results from shifting the factorial index incorrectly.

The relation between adjacent combinations is \(^{n}C_r=\frac{n-r+1}{r}\,^{n}C_{r-1}\). Substituting \(n=10\) and \(r=4\), the multiplier is \(\frac{10-4+1}{4}=\frac{7}{4}\). \(\frac{4}{7}\) is the reciprocal and would arise when expressing \(^{n}C_{r-1}\) in terms of \(^{n}C_r\). Exam tip: check which combination is being expressed in terms of the other before choosing the ratio.

Dividing the two factorial formulas gives the factor (\frac{n}{n-r}). In exams verify related identities using factorials.

The direct answer is option D, \(3!\). A permutation counts arrangements, while a combination only counts selections. After choosing 3 objects from n objects, the selected three can be arranged in \(3!=3\times2\times1=6\) orders. Therefore \(^{n}P_3=^{n}C_3\times3!\), so \(k=3!\). Option A, 3, is not enough because three selected objects have six possible orders, not three. Option B, \(^{n}C_3\), is the number of selections and would not provide the ordering factor. Option C, \(n-3\), is unrelated to arranging the selected objects. Option D is correct because the general identity is \(^{n}P_r=^{n}C_r r!\). Memory cue: combination chooses, factorial arranges.

The two combination terms are \(^{n}C_r\) and \(^{n-1}C_r\). In both terms, the subscript written with \(C\) is \(r\), so \(r\) is correct. \(n-1\) is the upper index of the second term, while \(n-r\) occurs in the denominator and is not a lower index. Exam tip: in \(^{n}C_r\), identify \(n\) as the upper index and \(r\) as the lower index.

Total selections are (2^6) and the empty selection is removed. In exams subtract (1) for non-empty selections.

After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.

The direct answer is option A: \(r+(r+2)=n\). A standard property says \(^{n}C_a=^{n}C_b\) when either \(a=b\) or \(a+b=n\). Here the lower indices are r and \(r+2\), and the question says they are different, so the equal-index case is excluded. Therefore they must be complementary: \(r+(r+2)=n\), or \(n=2r+2\). Option A states exactly this. Option B, \(r=0\) always, is not necessary and is not true for every valid equality. Option C, \(n=2r\) always, would describe a different relation and misses the extra 2. Option D, \(r+2=0\), is not the complement rule and would usually make the lower index invalid. Memory cue: equal combination values come from equal lower indices or complementary lower indices whose sum is n.

After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.

(^{n}C_2=\frac{n(n-1)}{2}) and (\frac{10\times9}{2}=45). In exams apply the pair-counting formula quickly.

The internal arrangements of three chosen objects are (3!). In exams remember (r!) while connecting (P) and (C).

(^{n}P_r=^{n}C_r\times r!), so the multiplier here is (4!). In exams count the arrangement of chosen objects separately.

Choosing (12) is like leaving (3), so (^{15}C_{12}=^{15}C_3). In exams reduce a large lower index using its complement.

In equal combinations, different lower indices are complementary, so (4+6=n). In exams check the sum of indices in equal combinations.

Put (n=9) and (r=5) in (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}). In exams the upper index of both terms decreases by (1).

Taking the ratio of factorial forms gives (^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}). In exams use ratio method for adjacent terms.

The ratio is (n-4+1=n-3), and (n-3=7) gives (n=10). In exams the ratio of consecutive permutations gives the last factor.

The (3!) orders of three chosen objects are counted extra in permutation. In exams divide by (r!) to get combination.

In (^{n}C_3=\frac{n!}{3!(n-3)!}), cancelling ((n-3)!) leaves (n(n-1)(n-2)). In exams write the cancellation step clearly.

(3!=6), and it removes internal orders of three chosen objects. In exams remember small factorial values.

A triangle needs an unordered selection of (3) points. In exams do not count the order of points while forming a shape.

First choose (3) candidates and then assign the posts in (3!) ways. In exams add arrangement when posts are different.