Permutations

Correct answer: A. 20

For each of the 5 snacks, any one of the 4 drinks can be selected. By the multiplication principle, the total number of ways is \(5 \times 4 = 20\). The value 9 results from addition, which is not appropriate for two successive independent choices. Exam tip: multiply the numbers of choices for independent selections made together.

Correct answer: C. 60

By the multiplication principle, when choices are made in successive independent stages, the total number of selections is the product of the choices at each stage. Thus, the total is \\(4 \\times 3 \\times 5 = 60\\). The value 45 results from multiplying only 3 and 5, so it omits the class choice. In exams, multiply the numbers of choices for all successive selections.

Correct answer: A. (60)

The choices for three places are (5,4,3), so (5\times4\times3=60). While filling positions, track remaining choices.

Correct answer: B. (120)

For three positions, choices are (6,5,4), so (6\times5\times4=120). Without repetition, a chosen letter is no longer available.

Correct answer: A. (120)

The direct answer is option A: 120 ways. There are 5 different books, and every book must be used once. For the first place, any of 5 books can be chosen. After that, 4 books remain for the second place, then 3, then 2, then 1. By the multiplication rule, the total is \(5\times4\times3\times2\times1=5!=120\). Option A is correct because it is exactly this factorial value. Option B, 25, does not count the successive choices and is not \(5!\). Option C, 20, is also not the product of all five decreasing choices. Option D, 10, is too small and does not represent arranging all five distinct books. Since changing the order changes the arrangement, this is a permutation question. Memory cue: arranging all different objects in a row means use \(n!\).

Correct answer: B. (120)

Here the posts are different, so order matters and the answer is ({}^{6}P_{3}=120). In exams, use permutation when posts are distinct.

Correct answer: C. (6)

The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.

Correct answer: D. 24

The governing idea is a permutation, because the flags are different and their order in a straight line matters. For the first position there are 4 choices, for the second there are 3 remaining choices, then 2, and finally 1. By the multiplication principle, the total number of arrangements is 4×3×2×1=4!=24. Thus option D is correct. This is not a combination problem, because combinations ignore order. The values 8, 12, and 16 do not account for all successive choices and therefore undercount the possible linear arrangements. Every distinct order, such as ABCD and BACD, is counted separately.

Correct answer: B. 5 पुस्तकों को एक शेल्फ पर क्रम से सजाना

When arranging 5 books on a shelf, changing their positions gives a different arrangement, so order matters; this is a permutation. In selecting a committee or team, order is irrelevant. Exam tip: identify a permutation whenever order matters.

Correct answer: B. (56)

Captain and vice-captain are distinct posts, so ({}^{8}P_{2}=56). When roles differ, order matters.

Correct answer: C. 6

Since all three available digits are distinct and none is zero, each permutation of 1, 2, and 3 forms a valid three-digit number. There are 3 choices for the hundreds place, 2 remaining choices for the tens place, and 1 choice for the units place. Hence the multiplication principle gives 3×2×1=3!=6. The six numbers are 123, 132, 213, 231, 312, and 321. Therefore option C is correct. The answer 3 counts only one position choice, while 9 incorrectly allows repetition through 3×3×3; 12 has no valid permutation basis. The phrase “without repetition” is essential because a digit cannot be used again after being placed.

Correct answer: D. (120)

The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.

Correct answer: A. 5

The governing concept is permutation notation. In general, nPr represents the number of ordered selections of r objects from n distinct objects, and nPr = n!/(n−r)!. Substituting n=5 and r=1 gives 5P1 = 5!/(5−1)! = 5!/4!. Since 5! = 5 × 4!, the 4! cancels and the result is 5. This also has a direct counting interpretation: one object can be selected from five distinct objects in five ways, and arranging one selected object adds no further choice. Therefore option A is correct. Option B confuses the result with a single arrangement or with 0!, while 10 and 25 do not follow the permutation formula.

Correct answer: B. (6)

The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.

Correct answer: C. (72)

Because the objects are selected and then placed in order, this is a permutation problem rather than a combination problem. For the first position, any of the 9 objects can be chosen. After one object has been used, 8 objects remain for the second position. By the multiplication principle, the total number of ordered arrangements is \(9\times8=72\), or \({}^{9}P_2=\frac{9!}{7!}=72\).

The order matters here: choosing object X first and object Y second is different from choosing Y first and X second. If order did not matter, the count would be \(\binom92=36\), which explains option B, but that is not the question being asked. Therefore the correct answer is 72, option C.

Correct answer: D. (42)

The direct answer is option D: 42 ways. First place and second place are different positions, so order matters. For first place there are 7 choices. After one runner is selected, 6 runners remain for second place. By the multiplication principle, the total is \(7\times6=42\), equivalently \({}^{7}P_2=\frac{7!}{(7-2)!}=42\). Option D is correct. Option A, 14, does not count all ordered pairs. Option B, 21, is \(\binom{7}{2}\), which counts pairs when order does not matter; it would be suitable for simply choosing two runners, not ranking them. Option C, 28, has no correct permutation interpretation here and is only half of 56. Since first and second are ranked, swapping the two runners creates a different result. Memory cue: positions, ranks, and arrangements usually mean permutation; use choices multiplied in order.

Correct answer: A. \({}^{8}P_{3}\)

Use permutations when order matters. Here, 3 books are selected and arranged from 8, so the expression is \({}^{8}P_{3}=\frac{8!}{5!}\). \({}^{8}C_{3}\) counts selection only, not arrangement. Exam tip: look for words such as “arrange” or “order.”

Correct answer: C. (60)

The prizes are distinct and students cannot repeat, so ({}^{5}P_{3}=60). Distribution of distinct prizes makes order important.

Correct answer: D. (24)

The word (MATH) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

Correct answer: A. 1

The standard permutation identity is nP0=1. From the formula nPr=n!/(n-r)!, putting r=0 gives nP0=n!/(n-0)!=n!/n!=1, provided n is a nonnegative integer. The combinatorial meaning agrees: selecting and arranging zero objects from n objects can be done in exactly one way, namely selecting nothing and leaving the arrangement empty. Therefore option A is correct. It is not 0, because there is one valid empty selection, and it is not n because no object is being selected. Although n!/n! also simplifies to 1, the requested numerical value is represented directly by option A.

Correct answer: B. (12)

Order matters on two special seats, so ({}^{4}P_{2}=12). If seats are distinct, use permutation.

Correct answer: C. (24)

The letters A, B, C, and D are four distinct choices. We need arrangements of 3 letters, and order matters: ABC and BAC are different arrangements. For the first position there are 4 choices. After one letter is used, 3 choices remain for the second position, and then 2 choices remain for the third. Therefore the total is \\(4\times3\times2=24\\), or \\({}^{4}P_3=24\\). Option C is correct.

This is a permutation, not a combination, because the question asks for ordered arrangements. Choosing only three letters without arranging them would use combinations and give a different count, but codes or arrangements distinguish position. Repetition is not indicated, and the four listed letters are distinct, so each choice is used at most once.

Correct answer: D. (5)

There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.

Correct answer: A. 840

This is a permutation because the order in which the pens are placed matters. Therefore, the number of arrangements is \({}^{7}P_{4}=\frac{7!}{(7-4)!}=7\times6\times5\times4=840\). The value 210 would result from an incorrect calculation and is not the required permutation. Exam tip: Use \({}^{n}P_r\) when both selection and order/position matter.

Correct answer: B. 6

The word SUN contains three distinct letters: S, U, and N. When all distinct letters are arranged in a row, the number of arrangements is the factorial of the number of letters. There are 3 choices for the first position, 2 for the second, and 1 for the last, so the total is 3×2×1=3!=6. The six arrangements are SUN, SNU, USN, UNS, NSU, and NUS. Therefore option B is correct. Repeated-letter adjustment is unnecessary because no letter occurs more than once. Option C counts only the letters, while 9 and 18 do not arise from the required factorial calculation.