Permutations

The (5) students receive one prize each from (7) prizes in an ordered way, so (^{7}P_5=2520). In exams, distinct recipients imply permutation.

The books are selected and arranged in order, so (^{9}P_4=3024). In exams, count order in shelf arrangements.

The speaking order is important, so (^{8}P_3=336). In exams, treat order of performance as permutation.

For divisibility by 5, the units digit must be 0 or 5. If the units digit is 0, there are 6 choices (1 to 6) for the thousands digit, and the two middle positions can be filled in \(^{5}P_2=20\) ways. This gives \(6\times20=120\) numbers. If the units digit is 5, the thousands digit cannot be 0, so it has 5 choices; the middle positions again have 20 arrangements. This gives \(5\times20=100\) numbers. Hence, the total is \(120+100=220\). Exam tip: when 0 is among the digits, always check that the leading digit is not 0.

One particular digit is required to occupy the third position, so that position has only one possible choice. The remaining five distinct digits must be arranged in the other five positions. The first of those positions can be filled in 5 ways, the next in 4 ways, and so on, giving \(5\times4\times3\times2\times1=5!=120\).

There is no extra choice for the fixed digit, because it is already specified. Thus we do not use \(6!\), which would incorrectly allow that digit to move. Also, since the problem says the digits are distinct and asks for six-digit numbers formed from them, the arrangement count is exactly the permutation of the remaining five digits. Hence the answer is 120, option A.

The specified object must remain in the last position, so that position is fixed and contributes only one possibility. The other eight distinct objects can be arranged freely in the first eight positions. Their number of arrangements is \(8!=8\times7\times6\times5\times4\times3\times2\times1=40320\).

We should not use \(9!\), because that would count arrangements in which the special object appears in positions other than the last. Fixing one position leaves a permutation of the remaining eight objects. Therefore the total number of valid arrangements is 40320, which is option B. The value 362880 in option A is \(9!\), so it ignores the restriction.

Subtract the cases where the two particular coins occupy the two ends, which are (2!\cdot6!), from total (8!). The answer is (40320-1440=38880).

The number of seating (7) students in (7) distinct positions is (7!=5040). In exams, use factorial when positions are distinct.

(6) of (8) children receive distinct toys, so (^{8}P_6=20160). In exams, distinct toys and receivers create a permutation.

The particular coin has (2) end choices and the remaining (10) coins are arranged in (10!) ways. The total is (2\cdot10!=7257600).

Only rotations are the same, so the circular arrangement is ((8-1)!=5040). In exams, separately check whether reflection is also the same.

The left-to-right order is important, so (^{7}P_5=2520). In exams, standing order means permutation.

The governing concept is permutation of distinct objects with one position already fixed. Because the first chapter is specified, that position has only one possible choice. The remaining five distinct chapters must be arranged in the five vacant positions, and their order matters. Thus the number of arrangements is 5! = 5 × 4 × 3 × 2 × 1 = 120. Therefore option A is correct. Option B, 720, equals 6! and would be appropriate only if no chapter were fixed. Options C and D are not the factorial count of all five remaining chapters; they omit some possible orders and therefore undercount the arrangements.

Changing the top-to-bottom order changes the design, so (^{9}P_4=3024). In exams, use permutation for ordered layers.

(6) questions are selected and attempted in order, so (^{8}P_6=20160). In exams, use permutation if the order of attempt differs.

Use the block method for permutations. Because all 5 mobiles must remain together, regard them as one single block. This mobile block plus the 4 distinct laptops gives 5 separate units, which can be arranged in 5! ways. Within the mobile block, the 5 distinct mobiles can be arranged among themselves in 5! ways. By the multiplication principle, the total is 5! × 5! = 120 × 120 = 14,400. Thus option B is correct. The other options result from using an incorrect number of units or omitting the internal arrangements of the mobile block; treating all nine objects as unrestricted would also violate the togetherness condition.

The calling order to the stage is important, so (^{10}P_4=5040). In exams, use permutation when order is specified.

Arrange Hindi books first in (6!) ways, then place (5) English books in (7) gaps in (^{7}P_5) ways. The total is (6!\cdot^{7}P_5=1814400).

There are (11) letters with (M,A,T) each repeated twice, so (\frac{11!}{2!2!2!}=4989600). In exams, divide by factorials of all repeated letters.

The relevant principle is permutation with repeated objects and a togetherness restriction. Combine the two Ns into one block, written NN. The objects to arrange are then NN, B, A, A, A: five objects in total, with three identical As. Their distinct arrangements are 5!/3! = (5 × 4 × 3 × 2 × 1)/(3 × 2 × 1) = 20. The NN block has no additional internal order because the two letters are identical. Thus option A is correct. Counting all letters without accounting for repeated As would overcount, while treating the two Ns as separate would include arrangements in which they are not adjacent.

Apply the block method. Since all four distinct science books must be consecutive, treat them as one science block. This block together with the six distinct mathematics books gives 7 units, which can be arranged in 7! ways. The four science books can be ordered inside their block in 4! ways. The mathematics books are not required to be together, so they remain separate units and no additional block restriction is imposed on them. Therefore the total is 7! × 4! = 5040 × 24 = 120,960. Option A is correct. Multiplying by 6! would incorrectly force the mathematics books into a block or count an extra internal arrangement already represented by their separate positions.

This is a permutation problem involving repeated letters. STATISTICS has 10 letters in total: S appears 3 times, T appears 3 times, I appears 2 times, and A and C appear once each. If every letter were treated as distinct, there would be 10! arrangements, but interchanging identical S, T, or I letters does not produce a new word arrangement. Therefore divide by the corresponding factorials: 10!/(3! × 3! × 2!) = 3,628,800/(6 × 6 × 2) = 3,628,800/72 = 50,400. Hence option A is correct. The other values result from ignoring one or more repeated-letter corrections and therefore overcount or miscount identical arrangements.

The governing rule is the permutation formula for objects with repetitions. ASSASSIN contains 8 letters: A occurs 2 times, S occurs 4 times, and I and N occur once each. Starting with 8! arrangements would temporarily treat equal letters as distinguishable. However, exchanging the two As changes nothing, so divide by 2!, and exchanging the four Ss also changes nothing, so divide by 4!. Thus the number of distinct arrangements is 8!/(2! × 4!) = 40,320/(2 × 24) = 40,320/48 = 840. Therefore option B is correct. Option A usually comes from forgetting the repeated As, while C and D use incorrect repetition adjustments.

The repeated-letter permutation formula is n!/(p!q!r!...) because rearrangements of identical letters do not produce new words. INDEPENDENCE has 12 letters: E occurs 4 times, N occurs 3 times, D occurs 2 times, and I, P, C occur once each. Thus the number of distinct arrangements is 12!/(4! × 3! × 2!). Compute 12! = 479,001,600 and 4! × 3! × 2! = 24 × 6 × 2 = 288. Therefore 479,001,600/288 = 1,663,200. Option A is correct. Ignoring any repeated group would produce an overcount, while dividing by an extra factor would undercount the valid arrangements.

For r = 2, the permutation formula is ⁿP₂ = n!/(n−2)! = n(n−1). Therefore n(n−1) = 90, which becomes n² − n − 90 = 0. Factoring gives (n−10)(n+9) = 0, so the algebraic possibilities are n = 10 and n = −9. Since n represents the number of available objects, it must be a positive integer; hence n = 10. Direct verification confirms that ¹⁰P₂ = 10×9 = 90. Thus option A is correct. The alternatives give 9×8 = 72, 11×10 = 110, and 12×11 = 132, so none of them satisfies the given equation. The negative root is rejected because a number of objects cannot be negative.