Permutations

Correct answer: C. 10

Using cancellation of factorials, \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Thus, \((n+5)(n+4)=210\). For \(n=10\), the product is \(15\times14=210\), so the correct answer is 10. Option 9 is incorrect because it gives \(14\times13=182\). Exam tip: cancel the common factorial terms before solving the resulting quadratic or product equation.

Correct answer: A. 57

\(\frac{21!}{19!\cdot2!}=\frac{21\times20}{2}=210\), and \(\frac{18!}{16!\cdot2!}=\frac{18\times17}{2}=153\). Therefore, the value is \(210-153=57\). Option 63 results from an incorrect evaluation of the second term as 147. Exam tip: cancel the factorials into a few consecutive factors before performing the arithmetic.

Correct answer: C. 12

Here, \(n\geq 2\). We have \(\frac{(n+1)!}{(n-1)!}=n(n+1)\) and \(\frac{n!}{(n-2)!}=n(n-1)\). Thus, \(n(n+1)-n(n-1)=2n\). Therefore, \(2n=24\), giving \(n=12\). Exam tip: Expand factorial quotients into products of consecutive integers before simplifying.

Correct answer: A. 210

\(\frac{22!}{20!\cdot 2!}=\frac{22\times21}{2}=231\), while \(\frac{21!}{20!}=21\). Therefore, the value is \(231-21=210\). The value 231 represents only the first term, not the complete expression. Exam tip: cancel common factorial factors before performing the arithmetic.

Correct answer: A. 2880

Use the block method together with circular permutations. Treat the four girls as one block because they must occupy consecutive seats. Along with the five individual boys, this creates 6 units around the circular table. The number of circular arrangements of these 6 units is (6 − 1)! = 5!, because rotations are identical. The four distinct girls can be arranged within their block in 4! ways. Therefore the required count is 5! × 4! = 120 × 24 = 2880, so option A is correct. The other options arise from using a linear formula, omitting internal girl arrangements, or overcounting rotations.

Correct answer: B. (2880)

Treat the (5) vowels as one block to get (4!\cdot5!=2880) arrangements. In exams, use the block method for together conditions.

Correct answer: A. (900)

First arrange the non-(R) letters in (\frac{5!}{2!}=60) ways and place two (R)'s in (\binom{6}{2}) gaps. The total is (60\cdot15=900).

Correct answer: B. (1260)

The last digit is (0) or (5), and both cases must be counted separately. The total is (^{7}P_4+6\cdot{}^{6}P_3=840+420=1260).

Correct answer: B. (3600)

Total circular arrangements are (7!), and together cases are (6!\cdot2!). Hence the answer is (5040-1440=3600).

Correct answer: A. (2880)

First seat the men around the circle in ((5-1)!) ways and then place the women in gaps in (5!) ways. The total is (4!\cdot5!=2880).

Correct answer: C. (360)

If the unit digit is (0), there are (6\cdot5\cdot4) ways, and if it is (2,4,6), there are (3\cdot5\cdot5\cdot4) ways. The total is (120+300=420).

Correct answer: B. (3720)

Subtract together cases (6!\cdot2!) and both-end cases (2!\cdot5!) from (7!). There is no overlap because both-end positions are not together, so the result is (5040-1440-240=3360).

Correct answer: A. (45360)

There are (9) letters with (M,T,E) each repeated twice, so (\frac{9!}{2!2!2!}=45360). In exams, put each repeated letter factorial in the denominator.

Correct answer: B. (6720)

Choose the position of the particular flag in (5) ways and fill the other (4) positions from (8) flags in (^{8}P_4) ways. The total is (5\cdot{}^{8}P_4=8400).

Correct answer: A. (900)

The first digit must be (5,6,7,8), and the last digit must be odd. Counting the cases gives (900) numbers.

Correct answer: A. (6720)

First choose (5) people in (\binom{8}{5}) ways and then seat them around a circle in ((5-1)!) ways. The total is (\binom{8}{5}\cdot4!=1344).

Correct answer: A. (265)

This is a derangement of (6) objects, whose number is (!6=265). In exams, when none is in the correct place, think derangement.

Correct answer: A. (6720)

Total valid (5)-digit numbers are (8\cdot{}^{8}P_4), and those without (0) are (^{8}P_5). The difference is (13440-6720=6720).

Correct answer: B. (1200)

Treat the two students as a block, count (6!\cdot2!) together arrangements and subtract cases where the third is at an end. The answer is (1200).

Correct answer: A. (720)

Treat the two beads as one block, so the necklace count for (7) units is (\frac{(7-1)!}{2}), with (2!) internal ways. The total is (720).

Correct answer: B. (1680)

The thousands place has (4) choices from (4,5,6,7), and the remaining places have (8\cdot7\cdot6) ways. The total is (1344).

Correct answer: A. (277200)

There are (11) letters with (E) three times, (N) three times, (G) twice and (I) twice. The number is (\frac{11!}{3!3!2!2!}=277200).

Correct answer: C. \({}^9P_4=3024\)

Excluding the particular object leaves 9 available objects. The number of ordered arrangements of 4 objects is \({}^9P_4=9\times8\times7\times6=3024\). Using \({}^{10}P_4\) would incorrectly allow the excluded object to be selected. Exam tip: for an “ordered arrangement,” use permutations \({}^nP_r\), not combinations.

Correct answer: A. (604800)

Arrange the men first in (6!) ways and place (4) women in (7) gaps in (^{7}P_4) ways. The total is (6!\cdot{}^{7}P_4=604800).

Correct answer: A. (4320)

The first digit can be (3,4,5,6,7), and the remaining (4) places are filled in (^{7}P_4) ways. The total is (5\cdot840=4200).