Permutations

Correct answer: D. (720)

The direct answer is option D, 720. There are 6 different posters and 6 positions from left to right. Choose a poster for the first position in 6 ways, then the second in 5 ways, followed by 4, 3, 2, and 1 ways. Hence the total is 6×5×4×3×2×1=6!=720. Option A, 120, is 5!, so it forgets one of the six posters. Option B, 36, does not represent the multiplication of decreasing choices and is not the required factorial. Option C, 60, is also not 6! and undercounts the arrangements. Option D is correct because every change in left-to-right order gives a new arrangement, and all posters are distinct. This is a linear arrangement, not a selection where order is ignored. Memory cue: arranging n distinct objects in a row gives n! ways; for six objects, calculate 6!=720.

Correct answer: A. (12)

The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.

Correct answer: B. 3024

Four files are to be selected from 9 distinct files and arranged in order, so permutations are used. The number of ways is \({}^{9}P_{4}=\frac{9!}{(9-4)!}=9\times8\times7\times6=3024\). Option 126 is \({}^{9}C_{4}\), which counts only selections and ignores order. Exam tip: words such as “arrange” or “in order” usually indicate \({}^{n}P_{r}\).

Correct answer: C. 30

RADAR has 5 letters, with 2 identical R's and 2 identical A's. Therefore, the number of distinct arrangements is \(\frac{5!}{2!\,2!}=\frac{120}{4}=30\). The value 120 would apply if all five letters were different; swapping identical letters does not create a new arrangement. Exam tip: for repeated objects, divide the total factorial by the factorial of each repeated group.

Correct answer: A. 40320

The governing idea is the permutation of distinct objects in a line. Since all 8 trophies are different and every trophy must be displayed, the first place has 8 choices, the next has 7, and the number of choices continues down to 1. Hence the total is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 8! = 40,320. Therefore, option A is correct. The number 5,040 equals 7!, so it omits one factor of 8. The number 720 equals 6!, and 64 is merely 8²; neither accounts for the successive choices created by arranging all eight distinct trophies. Because the trophies form a straight row, this is not a circular permutation.

Correct answer: B. 132

Although the wording says “chosen,” the two positions are different, so order matters. Selecting student X as class representative and student Y as assistant is different from selecting Y as class representative and X as assistant. There are 12 choices for the first post and, after that, 11 remaining choices for the second post. Thus the number of assignments is 12 × 11 = 132, or ¹²P₂ = 12!/(12−2)!. Therefore, option B is correct. The value 66 is ¹²C₂ and would be appropriate only if the two students formed an unlabelled group. The values 24 and 144 do not follow the required successive-choice calculation.

Correct answer: C. 120

The relevant concept is permutation of distinct letters. TRAIN has five letters—T, R, A, I, and N—and no letter is repeated. When all five letters are used, the first position can be selected in 5 ways, the second in 4 ways, and the remaining positions in 3, 2, and 1 ways. Therefore, the total number of different arrangements is 5 × 4 × 3 × 2 × 1 = 5! = 120. Hence option C is correct. No repeated-letter denominator is needed because every letter is distinct. The value 60 leaves out half of the arrangements, 25 is not the factorial count, and 10 does not represent the number of possible orders. The question asks for arrangements, so combinations would ignore order and be unsuitable.

Correct answer: D. (24)

For three places, there are (4\times3\times2=24) ways. Without repetition, each next choice decreases.

Correct answer: A. \({}^nP_r\)

Changing the order of the selected r objects creates a new arrangement, so it is a permutation: \({}^nP_r\). \({}^nC_r\) does not count order. Exam tip: choose P whenever order matters.

Correct answer: B. (210)

Playing order matters, so ({}^{7}P_{3}=7\times6\times5=210). Changing the order changes the list.

Correct answer: C. (60)

Since (P) occurs twice, arrangements are (\frac{5!}{2!}=60). Divide by the factorial of the repeated letter.

Correct answer: D. 30

The governing principle is an ordered selection of two students’ positions. Since the six desks are distinct, placing student A at desk 1 and student B at desk 2 is different from placing B at desk 1 and A at desk 2. Choose a desk for the first student in 6 ways; after that, choose a different desk for the second student in 5 ways. Thus the total is 6 × 5 = 30, which is ⁶P₂ = 6!/(6−2)!. Therefore, option D is correct. The value 12 counts too few arrangements, 15 would correspond to choosing two desks without assigning students to them, and 36 incorrectly allows the same desk to be used twice. Distinct seating places make order relevant.

Correct answer: A. 5040

The permutation formula is \,\({}^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \,\({}^{7}P_{7}=\frac{7!}{(7-7)!}=\frac{7!}{0!}=7!=5040\), since \,\(0!=1\). Option C, 720, is the value of \,\(6!\), not 7!. Exam tip: when \,\(r=n\), use \,\({}^{n}P_n=n!\) directly.

Correct answer: B. (16)

If the unit digit is (0), there are (4) ways, and if it is a non-zero even digit, there are (4\times3=12) ways, total (16). Make cases for zero questions.

Correct answer: C. (120)

(5) different kites can be distributed among (5) children in (5!=120) ways. Factorial appears in one-to-one distribution of distinct objects.

Correct answer: D. (504)

The solving order matters, so ({}^{9}P_{3}=9\times8\times7=504). Use permutation in ordered selection.

Correct answer: A. (6)

The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.

Correct answer: B. 156

The two offices have different titles and responsibilities, so they are distinguishable positions. Choosing candidate A as president and candidate B as vice-president is not the same as choosing B as president and A as vice-president. There are 13 choices for president. Once that choice is made, 12 candidates remain for vice-president. Therefore, the total is 13 × 12 = 156, or ¹³P₂ = 13!/(13−2)!. Hence option B is correct. The value 78 equals ¹³C₂ and would count only an unordered pair, ignoring which candidate receives which office. The values 26 and 169 do not represent the required two-stage selection without repetition.

Correct answer: D. (120)

The last position is fixed and the remaining (5) folders can be arranged in (5!=120) ways. Arrange the remaining items after fixing the item.

Correct answer: A. 14

The governing permutation formula is ⁿPᵣ = n!/(n−r)!. For r = 1, this becomes ⁿP₁ = n!/(n−1)! = n, because all factors from 1 to n−1 cancel. Substituting n = 14 gives ¹⁴P₁ = 14! / 13! = 14. Equivalently, arranging one object selected from 14 available objects simply means choosing which one object occupies the single position, so there are 14 possibilities. Therefore, option A is correct. The value 1 would ignore the 14 choices, 196 incorrectly squares 14, and 28 is twice 14 rather than the value of the permutation. No additional multiplication or division is needed when only one position is filled.

Correct answer: B. (120)

The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.

Correct answer: A. 5040

This is a permutation question because the order of the badges matters. Therefore, \({}^{10}P_{4}=10\times9\times8\times7=5040\). The value 210 is \({}^{10}C_{4}\), which counts only selections, not arrangements. Exam tip: use permutations whenever changing the order changes the outcome.

Correct answer: D. 20

Because the arrangement is ordered, this is a permutation rather than a combination. There are 5 choices for the first position. After one vowel has been used, 4 different vowels remain for the second position, so the total is 5 × 4 = 20. In formula form, ⁵P₂ = 5!/(5−2)! = 5!/3! = 20. Therefore, option D is correct. The value 10 would be ⁵C₂ and would count pairs without considering order; however, AE and EA are different ordered arrangements. The value 25 incorrectly allows the same letter to be used twice, and 5 counts only the first position. Since the question asks for two letters arranged in order and does not allow repetition, 20 is the unambiguous result.

Correct answer: A. (3)

The direct answer is option A, 3. The word DAD has 3 positions, but the two D letters are identical. If all three letters were different, there would be 3!=6 arrangements. However, interchanging the two D’s does not produce a visibly new word, so every distinct arrangement has been counted twice. Divide by 2!: 3!/2!=6/2=3. They are DAD, ADD, and DDA. Option A is correct. Option B, 6, incorrectly treats the repeated D letters as different. Option C, 2, misses one arrangement. Option D, 1, would be true only if there were no meaningful rearrangement, but three distinct placements are possible. The repeated-letter formula is n!/(r!), with one division for each repeated group. Memory cue: identical objects must not be labelled; divide by the factorial of their repetition count.

Correct answer: B. (6720)

The answer is ({}^{8}P_{5}=8\times7\times6\times5\times4=6720). Left-right order on a bench matters.