Derivations of formulas and their connections

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.

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

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.

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.

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

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}).

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.

(^{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.

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).

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

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

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

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

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.

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

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

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.

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.

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

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

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.

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

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

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

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