Permutations

Correct answer: B. (240)

The particular colour has (2) possible positions, and the remaining (3) positions are filled from (6) colours in (^{6}P_3) ways. The total is (2\cdot120=240).

Correct answer: A. (9979200)

The direct answer is option A: 9,979,200 distinct arrangements. The word has 11 letters. The letters B occur twice and I occur twice; the other seven letters occur once each. If all 11 positions were treated as different, there would be \(11!\) arrangements. However, swapping the two B's does not create a new visible arrangement, so divide by \(2!\); similarly divide by \(2!\) for the two I's. Therefore the answer is \(\frac{11!}{2!2!}=\frac{39,916,800}{4}=9,979,200\). Option A is correct. Option B is \(11!\), so it incorrectly treats repeated letters as distinct. Option C divides by too much or otherwise misses valid arrangements. Option D is exactly \(11!\), again ignoring repetitions. The important rule is: for repeated objects, divide by the factorial of each repetition count. Memory cue: distinct arrangements equal total factorial divided by repeated-letter factorials.

Correct answer: A. (28)

Choose the (2) incorrect letters in (\binom{8}{2}) ways and they must swap. Hence (\binom{8}{2}\cdot!2=28).

Correct answer: A. (720)

Choose the other (2) digits from (6) digits in (\binom{6}{2}) ways and arrange the (4) digits in (4!) ways. The total is (15\cdot24=360).

Correct answer: B. (6720)

The particular student has (3) inner positions, and the remaining (4) places are filled from (8) students in (^{8}P_4) ways. The total is (3\cdot1680=5040).

Correct answer: B. (3840)

Treat each couple as one block, giving (5!) block arrangements and (2!) ways inside each block. The total is (5!\cdot(2!)^5=3840).

Correct answer: A. (19958400)

The word contains eleven letters: P, E, R, M, U, T, A, T, I, O, and N. Every letter occurs once except T, which occurs twice. When repeated letters are present, simply using \\(11!\\) would count the same visible arrangement twice, because interchanging the two identical T letters does not create a new arrangement. We must therefore divide by \\(2!\\).

The number of distinct arrangements is \\(\\frac{11!}{2!}\\). Since \\(11!=39,916,800\\) and \\(2!=2\\), the result is \\(19,958,400\\). Thus option A is correct. The division is needed only for the repeated T letters; all other letters are distinct. This reasoning counts each different arrangement exactly once rather than counting identical copies separately.

Correct answer: A. (2880)

The two particular objects must have positions differing by (4), giving (4) position-pairs and (2!) orders. The remaining objects arrange in (6!) ways, so the total is (4\cdot2!\cdot6!=5760).

Correct answer: A. (6720)

There are (4) choices (1,3,5,7) for the last place, and the first place cannot be (0). The total is (4\cdot7\cdot{}^{7}P_3=5880).

Correct answer: A. (720)

Treat the three particular people as one block, giving (5) units around a circle with ((5-1)!) arrangements and (3!) internal ways. The total is (24\cdot6=144).

Correct answer: A. (4536)

Subtract the cases where the particular student is secretary, (^{9}P_3), from total (^{10}P_4). The answer is (5040-504=4536).

Correct answer: C. (360)

The first digit is fixed as (3), and the last place has (4) even choices. The middle (3) places are filled in (7\cdot6\cdot5) ways, giving (840).

Correct answer: A. (2880)

Choose exactly one of the two particular books in (\binom{2}{1}) ways and the other (4) books from (6). Then arrange (5!) ways, giving (2\cdot\binom{6}{4}\cdot5!=3600).

Correct answer: A. (720)

The two empty chairs are fixed, so seat (6) persons on the remaining (6) chairs in (6!) ways. The answer is (720).

Correct answer: A. (5040)

Treat the three objects as a fixed-order block, giving (7) units. Hence there are (7!=5040) arrangements.

Correct answer: A. (1344)

The thousands place has (3) choices (1,2,3), and the remaining (3) places have (8\cdot7\cdot6) ways. The total is (1008).

Correct answer: A. (384)

Treat each couple as one block, so (4) blocks have ((4-1)!) circular arrangements and ((2!)^4) internal ways. The total is (96).

Correct answer: B. (1260)

The particular student has (3) non-first positions, and the other (3) positions are filled from (7) students in (^{7}P_3) ways. The total is (3\cdot210=630).

Correct answer: B. (1800)

Treat (1) and (8) as a block and choose the other (3) digits from (6). The total is (\binom{6}{3}\cdot4!\cdot2!=960).

Correct answer: A. (322560)

Subtract cases where the three particular books are together, (7!\cdot3!), from total (9!). The answer is (362880-30240=332640).

Correct answer: A. (21)

Choose the (2) incorrect letters in (\binom{7}{2}) ways, and they derange in only (1) way. The answer is (21).

Correct answer: A. (1260)

Treat the four (S)'s as one block, giving (8) units with (I) four times and (P) twice. The number is (\frac{8!}{4!2!}=840).

Correct answer: A. (4320)

Total circular arrangements are (7!). If the particular object has the two given objects as neighbours, they can be placed around it in (2!) ways and the rest in (5!) ways. The answer is (5040-240=4800).

Correct answer: B. (67200)

There are (5) odd choices for the first digit and (5) even choices for the last digit. The middle (4) places are filled from (8) digits in (^{8}P_4) ways, giving (5\cdot5\cdot1680=42000).

Correct answer: C. (10080)

Out of total (8!) arrangements, each independent order condition keeps half the arrangements. Hence the answer is (\frac{8!}{2\cdot2}=10080).