Derivations of formulas and their connections

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

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.

In equal combinations, different lower indices are complementary, so (5+8=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=6) gives (\frac{n-5}{6}).

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.

(^{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.

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).

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

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

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

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

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.

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

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

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.

(^{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).

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

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

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.

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

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

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

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

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.

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.