Derivations of formulas and their connections

Correct answer: A. (^{n}C_r) counts selection and (^{n}P_r) counts ordered selection

Explanation: Permutation adds the arrangement of chosen objects to combination. In exams remember (P=C\times r!) conceptually.

Correct answer: C. (n-r+1)

Explanation: In the permutation recurrence, the new factor is the choices for the (r)th position, (n-r+1). In exams connect the multiplier with the next position.

Correct answer: A. (13)

Explanation: In equal combinations, different lower indices are complementary, so (5+8=n). In exams check the sum of indices in equal (C) terms.

Correct answer: B. (\frac{n-5}{6})

Explanation: The general ratio is (\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}). Putting (r=6) gives (\frac{n-5}{6}).

Correct answer: C. (14)

Explanation: The ratio is (n-6+1=n-5), and (n-5=9) gives (n=14). In exams the ratio of consecutive permutations gives the last factor.

Correct answer: B. (^{14}C_{11}=\frac{^{14}P_3}{3!})

Explanation: (^{14}C_{11}=^{14}C_3), and (^{14}C_3=\frac{^{14}P_3}{3!}). In exams first use the complement to get a smaller index.

Correct answer: D. (^{10}C_7+^{10}C_6)

Explanation: Put (n=11) and (r=7) 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. (^{n-1}C_r)

Explanation: In Pascal's identity, the other case is not choosing the special object, (^{n-1}C_r). In exams separate included and excluded cases.

Correct answer: C. Both are equal

Explanation: The coefficients are (^{10}C_3) and (^{10}C_7), and (3+7=10). In exams coefficients of complementary powers are equal.

Correct answer: A. Because (5+6=11)

Explanation: The coefficients are (^{11}C_5) and (^{11}C_6), which have complementary indices. In exams identify binomial symmetry.

Correct answer: C. (2^{n-1})

Explanation: The sum of odd indexed combinations equals the even indexed sum, (2^{n-1}). In exams remember the alternating identity from ((1-1)^n).

Correct answer: A. \(n-5\)

Explanation: The product form of a permutation is \({}^{n}P_r=n(n-1)(n-2)\cdots(n-r+1)\). Hence, its last factor is \(n-r+1\). On putting \(r=6\), we get \(n-6+1=n-5\), so option A is correct. \(n-6\) misses the required \(+1\). Exam tip: the product must contain exactly \(r\) factors; from \(n\) to \(n-5\), there are 6 factors.

Correct answer: C. (6)

Explanation: There are (r=6) positions to fill, so there are (6) factors. In exams connect the number of factors with positions.

Correct answer: D. (15)

Explanation: (^{n}C_2=\frac{n(n-1)}{2}), and (\frac{15\times14}{2}=105). In exams identify (n) using the pair formula.

Correct answer: B. (15)

Explanation: The direct answer is option B, \(n=15\). For two ordered selections, \(^{n}P_2=\frac{n!}{(n-2)!}=n(n-1)\). The given equation therefore becomes \(n(n-1)=210\). We need two consecutive numbers whose product is 210. Since \(15\times14=210\), \(n=15\). Algebraically, \(n^2-n-210=0\), which factors as \((n-15)(n+14)=0\). The possible roots are 15 and -14, but n counts objects and must be positive, so n=15. Option A, 14, gives \(14\times13=182\). Option B, 15, gives 210 and is correct. Option C, 16, gives 240. Option D, 13, gives 156. Do not divide 210 by 2: permutations count ordered pairs, so AB and BA are different.

Correct answer: C. (5!)

Explanation: (^{n}C_5=\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}). In exams keep (r!) in the denominator of (^{n}C_r).

Correct answer: C. (5)

Explanation: After cancelling ((n-5)!), there are (5) factors from (n) to (n-4). In exams the number of numerator factors is (r).

Correct answer: C. (^{10}C_5\times5)

Explanation: First there are (^{10}C_5) ways to form the committee and (5) choices for coordinator. In exams multiply when a role follows selection.

Correct answer: A. (^{10}P_5\div4!)

Explanation: One person has a distinct role and the order of the remaining (4) members is irrelevant. In exams remove overcounting of remaining members when there is a special role.

Correct answer: D. (^{7}C_3\times{}^{6}C_2)

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

Correct answer: C. (^{6}C_3{}^{7}C_2+^{6}C_4{}^{7}C_1+^{6}C_5{}^{7}C_0)

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

Correct answer: A. (^{9}C_6\times6!)

Explanation: First choose (6) letters and then arrange them in (6!) ways. In exams order is important in a word.

Correct answer: D. (9^6)

Explanation: Each position has (9) choices available again. In exams use the power rule (n^r) when repetition is allowed.

Correct answer: A. (7\times8^4)

Explanation: The first digit cannot be (0), so there are (7) choices and (8) choices for each of the remaining four places. In exams treat leading zero separately.

Correct answer: B. (7\times{}^{7}P_4)

Explanation: There are (7) non-zero choices for the first place, and then (4) places are filled from the remaining (7) digits. In exams handle the first place separately.