Derivations of formulas and their connections

Correct answer: C. (n-r)

Explanation: 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}).

Correct answer: A. (9\times8\times7)

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

Correct answer: B. (n-2)

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

Correct answer: B. (\frac{n(n-1)(n-2)}{6})

Explanation: 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).

Correct answer: A. Because each pair is counted in (2!) orders

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

Correct answer: B. (^{n}P_3=^{n}C_3\times3!)

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

Correct answer: C. (^{n}C_3=\frac{^{n}P_3}{3!})

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

Correct answer: A. Because choosing (n-1) is like leaving one object

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

Correct answer: A. जब केवल एक वस्तु चुनी जाए

Explanation: 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.

Correct answer: A. The number of ways to arrange the selected \(r\) objects

Explanation: \(^{n}C_r\) 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.

Correct answer: C. (^{12}C_3)

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

Correct answer: A. By including or excluding one special object

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

Correct answer: A. (^{6}C_3+^{6}C_2)

Explanation: 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.

Correct answer: A. Because choosing r objects from n objects is impossible

Explanation: 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.

Correct answer: B. Because more than (n) positions cannot be filled without repetition

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

Correct answer: A. Because there are (^{n}C_r) ways to choose the second term (r) times

Explanation: 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.

Correct answer: C. (^{n}C_r)

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

Correct answer: A. (^{n}C_0+^{n}C_1+\cdots+^{n}C_n=2^n)

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

Correct answer: A. Difference between even indexed and odd indexed coefficients is (0)

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

Correct answer: A. Each object has two options include or exclude

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

Correct answer: B. (2^{10}-1)

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

Correct answer: A. Because interchanging identical letters does not make a new arrangement

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

Correct answer: C. (\frac{7!}{3!})

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

Correct answer: B. ((5-1)!)

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

Correct answer: C. (4^3)

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