Permutations

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.

\(\frac{21!}{19!\cdot2!}=\frac{21\times20}{2}=210\), 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.

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.

\(\frac{22!}{20!\cdot 2!}=\frac{22\times21}{2}=231\), 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.

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.

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

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).

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).

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

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).

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).

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).

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.

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).

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

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).

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

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).

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).

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).

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

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).

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.

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).

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).