Derivations of formulas and their connections

Adding the remaining factor ((n-r)!) makes the numerator (n!). In exams identify the missing tail while forming factorial form.

In combination the order of selected objects is not considered different. In exams remove internal order in unordered selection.

The governing identity is the symmetry, or complementary-selection, property of combinations: ⁿCᵣ = ⁿCₙ₋ᵣ. It says that choosing r objects from n is equivalent to choosing the n-r objects that are left out. In this question, n = 13 and r = 10, so the complementary index is 13 - 10 = 3. Hence ¹³C₁₀ = ¹³C₃, making option B the required identity. Option A relates permutations and combinations through r!, but it does not directly replace the lower index. Option C is Pascal’s identity, which expresses a combination as a sum, and option D is the permutation formula; neither gives this immediate conversion.

First there are (^{n}C_r) selections and then (r!) arrangements for each selection. In exams multiply for choose then arrange.

For four ordered positions the choices are (9), (8), (7), (6). In exams each next choice decreases by one without repetition.

Simplifying the factorial form gives the factor (\frac{n}{r}). In exams check adjacent upper index identities by ratio.

The direct answer is option C: the lower index that remains unchanged is r. In the displayed relation, the two combinations are \(^{n}C_r\) and \(^{n-1}C_r\). The upper index changes from n to n−1, but the lower index is r in both. Option A, r−1, is not written in either combination. Option B, n−r, is a quantity appearing in factorial formulas but is not the lower index shown here. Option D, n−1, is the new upper index, not a lower index. Read combination notation carefully: in \(^{n}C_r\), n is the upper number and r is the lower number. Therefore C is certain. A useful check is to compare the position of the small subscript below C rather than focusing only on the symbols around it.

Put (n=8) and (r=4) in (^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}). In exams the upper index of both terms decreases by (1).

There are (^{n}C_r) ways to choose (b) from (r) of the (n) brackets. In exams treat a binomial coefficient as selection.

We choose (x) twice from (7) brackets. In exams the coefficient of (x^r) in ((1+x)^n) is (^{n}C_r).

One group has (r) chosen objects and the other has (n-r) remaining objects. In exams check that group sizes add to (n).

Once the committee of (5) is chosen the remaining (7) students are fixed. In exams use only one combination when the complement group is fixed automatically.

First choose the books and then assign display order. In exams multiply by (r!) when order follows selection.

There are no different posts among the representatives so order is not counted. In exams use combination for only selection.

For two ordered positions the choices are (n) and (n-1). In exams treat (^{n}P_2) as the ordered pair formula.

In a combination, the order of the chosen \(r\) objects does not matter, but it matters in a permutation. Each combination can be arranged in \(r!\) orders; hence \(^{n}P_r=r!\,{}^{n}C_r\). Exam tip: use the \(r!\) factor whenever order is counted.

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.

Dividing the factorial formulas gives (\frac{n-r}{r+1}). In exams ratios make adjacent combinations faster.

In equal combinations the lower indices are equal or complementary. Here the indices are different so (r+(r+2)=n).

(^{n}C_n=^{n}C_0) and both have value (1). In exams remember both all selected and none selected boundary cases.

If the special object is selected the remaining (r-1) objects are chosen from (n-1). In exams the lower index decreases by (1) in the included case.

After removing the special object all (r) objects are chosen from (n-1). In exams the lower index remains the same in the excluded case.

Including or excluding one fixed object gives (^{9}C_3) and (^{9}C_4). In exams the upper index decreases by (1) in a Pascal split.

Each position has (4) choices again and again. In exams use the power formula (n^r) when repetition is allowed.

For an ordered code without repetition there are (4\times3\times2) ways. In exams a code has order.