Derivations of formulas and their connections

Correct answer: B. ((n-r)!)

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

Correct answer: A. Internal orders of selected (r) objects

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

Correct answer: B. ⁿCᵣ = ⁿCₙ₋ᵣ

Explanation: 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.

Correct answer: C. (^{n}C_r\times r!)

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

Correct answer: A. (9\times8\times7\times6)

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

Correct answer: B. (^{n}C_r=\frac{n}{r},^{n-1}C_{r-1})

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

Correct answer: C. (r)

Explanation: 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.

Correct answer: A. (^{7}C_4+^{7}C_3)

Explanation: 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).

Correct answer: A. Choosing (b) from (r) brackets

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

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

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

Correct answer: C. (r) और (n-r)

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

Correct answer: A. (^{12}C_5)

Explanation: 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.

Correct answer: B. (^{5}C_2\times2!)

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

Correct answer: B. (^{7}C_3)

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

Correct answer: A. (n(n-1))

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

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

Explanation: 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.

Correct answer: A. Choices for the (r)th position

Explanation: 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.

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

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

Correct answer: A. (r+r+2=n)

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

Correct answer: B. (^{n}C_0)

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

Correct answer: B. (^{n-1}C_{r-1})

Explanation: 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.

Correct answer: A. (^{n-1}C_r)

Explanation: 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.

Correct answer: A. (^{9}C_4+^{9}C_3)

Explanation: 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.

Correct answer: C. (4^3)

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

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

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