Permutations

Permutations apply when order or roles matter. President, secretary, and treasurer are distinct posts, so interchanging the same 3 students gives a different arrangement. Mere team or object selection uses combinations. Exam tip: named positions usually indicate permutations.

The direct answer is option B: 336 ways. We must choose and order 3 books from 8 different books. The first position has 8 choices, the second has 7 choices because repetition is forbidden, and the third has 6 choices. Therefore the number of ordered selections is \(8\times7\times6=336\), or \({}^{8}P_3=\frac{8!}{5!}\). Option B is correct. Option A, 120, is not the required product and would correspond to a different arrangement count. Option C, 56, is \(\binom{8}{3}\), which selects three books without considering their order; here the books are placed in order, so it is insufficient. Option D, 512, incorrectly uses \(8^3\), which would allow repetition. Exam cue: “place,” “arrange,” or “in order” usually means order matters.

Since the order of covers on the shelf matters, use permutations. There are 6 choices for the first position, 5 for the second, and 4 for the third. Thus, the total number of arrangements is \({}^{6}P_{3}=6\times5\times4=120\). The value 60 is \({}^{6}C_{3}\), which applies when order does not matter. Exam tip: Use permutations for ordered placements in a row or on a shelf.

There are (4) choices for the first place and (3) for the second, so (4\times3=12). In passwords, changing positions changes the arrangement.

The governing concept is permutation in a linear arrangement. Since all seven certificates are distinct and every certificate must be used, the order of the certificates matters. For the first position there are 7 choices, for the second 6 remaining choices, then 5, 4, 3, 2, and 1. By the multiplication principle, the total number of arrangements is 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7! = 5040. Therefore, option A is correct. Option B is 6!, so it effectively omits one object. Option D is 5!, and option C is only 7², neither of which counts all possible linear orders. A combination formula is inappropriate because the certificates are being arranged, not merely selected.

First and second positions are different positions, so assigning one student first and another second is an ordered selection. There are 9 choices for first place. Once that student is selected, 8 students remain for second place. Thus the multiplication principle gives 9 × 8 = 72, or 9P2 = 9!/(9−2)! = 72. Option B is therefore correct. If order did not matter, 9C2 = 36 would be used, which explains the tempting option C. The values 18 and 81 do not account correctly for the two successive, non-repeating position choices.

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

There are (4) choices for tens and (3) for units, so (4\times3=12). Changing places changes the number.

When order matters, the same selected objects in different orders form different arrangements, so \({}^nP_r=\frac{n!}{(n-r)!}\). Option B is for combinations, where order is ignored. Exam tip: choose permutation whenever order matters.

There are (6) choices for the first position and (5) for the second, so (30). Order matters in distinct positions.

The direct answer is C: 60. APPLE has five positions, but P appears twice; the two P letters are identical and must not be counted as different. If temporarily treated as distinct, there are \(5!=120\) orders. Every actual arrangement is then counted \(2!\) times by exchanging the two P's, so the number is \(5!/2!=120/2=60\). Option C is correct. Option A, 120, overcounts by treating identical P's as different. Option B, 20, is too small and has no valid counting basis here. Option D, 30, also does not follow from the repetition formula. The useful rule is \(n!/(r_1!r_2!\cdots)\) for repeated letters.

The chairs are distinct, so ({}^{5}P_{3}=5\times4\times3=60). Changing seats changes the arrangement.

The permutation formula is \({}^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \({}^{6}P_{6}=\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!=720\), since \(0!=1\). The option 120 is the value of \(5!\), so it is not correct. Exam tip: when \(r=n\), use \({}^{n}P_n=n!\) directly.

Zero cannot be in the tens place, so (3\times3=9) numbers are possible. Check the first-place condition in number formation.

(4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.

The three positions are distinct, so ({}^{7}P_{3}=7\times6\times5=210). Use permutation when positions have different names.

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

The three posts are distinct, so ({}^{10}P_{3}=10\times9\times8=720). Order matters for different posts.

The fundamental principle of counting applies because a two-digit number is formed by making one choice for the tens place and one choice for the units place. Each place can contain 2, 5 or 8, so there are 3 choices for the tens digit and, since repetition is allowed, 3 independent choices for the units digit. Therefore the total is 3 × 3 = 9. Option C is correct. The value 6 would apply if repetition were forbidden, because then the choices would be 3 × 2. Option D counts only one position, while 12 has no valid product for the stated conditions.

The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.

The governing concept is permutation notation. The expression nPr counts ordered selections of r objects from n distinct objects, and its formula is nPr = n!/(n−r)!. For r = 1, one position is being filled, so any one of the nine objects may occupy it. Thus 9P1 = 9!/(9−1)! = 9!/8! = 9. Therefore, option A is correct. The same result follows directly from the special identity nP1 = n. Option B may result from confusing one selected object with one possible arrangement, while option C incorrectly calculates 9 × 9. Option D has no support in the permutation formula. Since only one object is ordered, no additional multiplication by 9 is required.

The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.

The governing concept is permutation because four notebooks are chosen from eight distinct notebooks and then placed in a specified order. The first position can contain any of 8 notebooks, the second any of the remaining 7, the third any of the remaining 6, and the fourth any of the remaining 5. Hence the multiplication principle gives 8 × 7 × 6 × 5 = 1680. Using notation, this is 8P4 = 8!/(8−4)! = 8!/4! = 1680. Therefore, option A is correct. Option D, 70, equals 8C4 and counts selection without order. Option C is 6!, and option B does not equal the required product. Because the table positions are ordered, combinations alone cannot be used.

The direct answer is option B, 60. We must arrange 3 letters selected from 5 distinct letters A, B, C, D, and E, without repetition. For the first position there are 5 choices. After choosing one letter, 4 choices remain for the second position. Then 3 choices remain for the third position. Thus the total is 5×4×3=60, or ^5P_3=5!/(5−3)!=60. Option A, 10, counts selections in an unsuitable way and does not account for order. Option B is correct because both selection and arrangement matter. Option C, 15, is not the number of ordered 3-letter arrangements; it is also not ^5C_3, which equals 10. Option D, 125, would come from 5 choices at every position, which would allow repetition, but repetition is not stated as allowed here. For example, ABC and BAC are different arrangements. Exam cue: “order” means permutation, and without repetition the choices decrease: 5, 4, 3.

(M) occurs twice, so there are (\frac{3!}{2!}=3) arrangements. Repeated letters reduce total arrangements.