Derivations of formulas and their connections

Chosen and not chosen objects together make (n) so the remaining number is (n-r). In exams this idea connects to (^{n}C_r=^{n}C_{n-r}).

This is ordered selection so the choices are (9) then (8) then (7). In exams treat position based questions as order important.

For three positions the factors are (n), (n-1), (n-2). In exams use (n-r+1) for the last factor.

From (^{n}C_3=\frac{n!}{3!(n-3)!}) the numerator becomes (n(n-1)(n-2)) and denominator (6). In exams remember (3!=6).

The pair (AB) and (BA) is considered same so we divide by (2!). In exams remove overcounting when unordered appears.

First choose (3) objects and arrange them in (3!) ways. In exams write ordered selection as selection times arrangement.

Permutation counts (3!) internal orders extra. In exams divide by these orders to get combination.

Choosing (n-1) objects means leaving (1) object and there are (n) choices for that. In exams think by complementary selection.

When only one object is chosen, order cannot make a difference. Here \(^{n}P_1=\frac{n!}{(n-1)!}=n\) and \(^{n}C_1=n\). Exam tip: for a single selection, arrangement is irrelevant.

\(^{n}C_r\) counts only selections, whereas permutations also consider order. The selected \(r\) objects can be ordered in \(r!\) ways; for example, when \(r=3\), \(3!=6\). Exam tip: check whether order matters.

Choosing (9) is same as leaving (3) so (^{12}C_9=^{12}C_3). In exams replace a large index by (n-r).

The total selection is split into two cases according to whether one special object appears or not. In exams remember the fixed object method.

The direct answer is option A: \(^{7}C_3=^{6}C_3+^{6}C_2\). Pascal’s relation is \(^{n}C_r=^{n-1}C_{r-1}+^{n-1}C_r\). Substitute \(n=7\) and \(r=3\): the first lower index becomes \(3-1=2\), and the second remains 3; the upper index becomes 6 in both terms. Hence \(^{7}C_3=^{6}C_2+^{6}C_3\), and addition can be written in either order, exactly matching A. Option B incorrectly leaves one term with upper index 7. Option C uses \(^{6}C_4\), which is not the direct Pascal substitution, although symmetry could make related expressions equal in another argument. Option D adds terms with upper index 7 and is not Pascal’s relation. Reduce the upper index by one and change only one lower index by one.

The governing counting concept is that nCr represents the number of ways to choose an r-element subset from n available objects, with order ignored. When r > n, the requested selection needs more objects than are available, so not even one valid selection can be formed. Consequently, the number of such selections is defined as 0. Option A states this direct combinatorial reason. Option B describes a possible distinction between permutations and combinations but is irrelevant here; order never becomes important merely because r exceeds n. Option C has no connection with the condition. Option D is false because the standard factorial convention is 0! = 1, not 0. Thus option A is the only correct explanation.

Without repetition more than (n) positions cannot be filled using (n) distinct objects. In exams check the availability condition first.

In ((a+b)^n) the ways to choose (b) from (r) of the (n) brackets are (^{n}C_r). In exams connect coefficient with selection.

There are (^{n}C_r) ways to choose (x) from (r) brackets. In exams the coefficient in ((1+x)^n) is directly a combination.

In binomial expansion the sum of all coefficients becomes (2^n). In exams use the same identity for total subsets.

From ((1-1)^n=0) the even and odd combination sums are equal. In exams connect alternating sum with (0).

Each object can independently be included or excluded. In exams think of subset questions as two choices per object.

Total selections are (2^{10}) and the empty selection must be removed. In exams subtract the empty case for at least one.

Internal arrangements of identical letters are not visibly different. In exams identify repeated letters and divide by their factorials.

In total (7!) arrangements the (3!) internal orders of (A) are identical. In exams put the factorial of the repeated object in the denominator.

Rotations are same in a circle so fixing one person gives (4!) ways. In exams use the one fixed rule in circular seating.

Each position again has (4) choices so the number of ways is (4^3). In exams do not decrease options when repetition is allowed.