Derivations of formulas and their connections

Correct answer: A. Because an empty selection and an empty arrangement both have count 1

Explanation: When \(r=0\), there is exactly one way to select no object: select nothing. Similarly, there is exactly one way to arrange zero objects. Using the formulas, \(^{n}P_0=\frac{n!}{n!}=1\) and \(^{n}C_0=\frac{n!}{0!n!}=1\), since \(0!=1\). Option D is incorrect because \(^{n}C_0=1\), not n. Exam tip: remember that an empty selection or an empty arrangement has count 1.

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

Explanation: Combination is obtained by removing (r!) internal orders counted in permutation. In exams divide when going from ordered to unordered.

Correct answer: B. (4)

Explanation: The relation is (^{n}P_r=^{n}C_r r!), and (20) is not equal to any (r!) for these options. With these options the question would be invalid.

Correct answer: C. 4

Explanation: The governing connection is ⁿPᵣ = ⁿCᵣ × r!. A selected group of r objects can be arranged in r! different orders, which explains this identity. Comparing it with the given relation ⁿPᵣ = 24 × ⁿCᵣ and cancelling the common nonzero factor ⁿCᵣ gives r! = 24. Since 4! = 4 × 3 × 2 × 1 = 24, the value is r = 4, so option C is correct. The other choices do not work: 2! = 2, 3! = 6, and 5! = 120. Notice that n is unnecessary because the ratio ⁿPᵣ/ⁿCᵣ is always r!, so the information directly determines r.

Correct answer: B. (10)

Explanation: In equal combinations different lower indices are complementary, so (3+7=n). In exams check the sum of indices in equal (C) terms.

Correct answer: A. (\frac{n-4}{5})

Explanation: The general ratio is (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}). Putting (r=5) gives (\frac{n-4}{5}).

Correct answer: B. (10)

Explanation: The ratio is (n-5+1=n-4), and (n-4=6) gives (n=10). In exams the ratio of consecutive permutations gives the last factor.

Correct answer: A. (^{11}C_8=\frac{^{11}P_3}{3!})

Explanation: (^{11}C_8=^{11}C_3) and (^{11}C_3=\frac{^{11}P_3}{3!}). In exams first use complement to get a smaller index.

Correct answer: A. (^{9}C_6+^{9}C_5)

Explanation: Put (n=10) and (r=6) 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: C. (^{n-1}C_{r-1})

Explanation: In Pascal's identity the second term is (^{n-1}C_{r-1}). In exams remember the included case of one special object.

Correct answer: A. Because (4+5=9)

Explanation: The coefficients are (^{9}C_4) and (^{9}C_5), which have complementary indices. In exams check symmetry in binomial coefficients.

Correct answer: A. Both are equal

Explanation: The coefficients are (^{8}C_3) and (^{8}C_5), and (3+5=8). In exams coefficients of complementary powers are equal.

Correct answer: A. Because even and odd indexed sums are equal

Explanation: From ((1+1)^n) and ((1-1)^n), even and odd sums are equal. In exams remember the alternating sum identity.

Correct answer: B. \(n-4\)

Explanation: The product form of a permutation is \(^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\). Hence, the last factor is \(n-r+1\). Substituting \(r=5\) gives \(n-5+1=n-4\). \(n-5\) is a common near-miss because the \(+1\) in the last factor is omitted. Exam tip: count the \(r\) factors in the product form to verify the final factor.

Correct answer: C. (5)

Explanation: There are (r=5) positions to fill, so there will be (5) factors. In exams connect the number of factors with selected positions.

Correct answer: C. (13)

Explanation: (^{n}C_2=\frac{n(n-1)}{2}) and (\frac{13\times12}{2}=78). In exams identify (n) using the pair formula.

Correct answer: A. \({}^nP_r = {}^nC_r \times r!\)

Explanation: In a combination, order is ignored, but in a permutation it matters. Each selection of \(r\) objects can be arranged in \(r!\) ways; hence \({}^nP_r={}^nC_r\times r!\). Exam tip: first check whether order matters.

Correct answer: C. (4!)

Explanation: The direct answer is option C, \(4!\). Start with \(^{n}C_r=\frac{n!}{r!(n-r)!}\). For r=4, \(^{n}C_4=\frac{n!}{4!(n-4)!}\). Expanding and cancelling \((n-4)!\) gives \(^{n}C_4=\frac{n(n-1)(n-2)(n-3)}{4!}\). Thus the simplified denominator is \(4!=24\). Option A, \(2!\), is too small and would correspond to two selected objects. Option B, \(3!\), would correspond to three selected objects. Option C is correct because the lower index 4 becomes the factorial in the denominator. Option D, \(n!\), belongs to the original factorial formula but is cancelled during simplification. Memory cue: in \(^{n}C_r\), the selected-number factorial \(r!\) stays in the denominator.

Correct answer: A. (n(n-1)(n-2)(n-3))

Explanation: After cancelling ((n-4)!), four factors from (n) to (n-3) remain. In exams the number of numerator factors is (r).

Correct answer: A. (^{9}C_4\times4)

Explanation: First there are (^{9}C_4) ways for the committee and (4) choices for leader. In exams multiply when a role follows selection.

Correct answer: A. (^{9}P_4\div3!)

Explanation: One leader is distinct and the order of the other (3) members is irrelevant, so divide (^{9}P_4) by (3!). In exams separate roles and ordinary members.

Correct answer: C. (^{6}C_3\times{}^{5}C_2)

Explanation: There is unordered selection in two categories, so combinations are multiplied. In exams count category-wise selections separately.

Correct answer: A. (^{6}C_3{}^{5}C_2+^{6}C_4{}^{5}C_1+^{6}C_5{}^{5}C_0)

Explanation: The number of boys can be (3), (4), or (5). In exams add all valid cases in at least conditions.

Correct answer: A. (^{8}C_5\times5!)

Explanation: First choose (5) letters and then arrange them in (5!) ways. In exams order is important in a word.

Correct answer: C. (8^5)

Explanation: Each position has (8) choices available again. In exams use the power rule (n^r) when repetition is allowed.