math-eval/TAL-SCQ5K huggingface dataset:
Problem: There are  different ways for a librarian, lending six books to three students, given that each student gets a book. 
Options:
aoVal: A
content: $$120$$ 
aoVal: B
content: $$100$$ 
aoVal: C
content: $$96$$ 
aoVal: D
content: $$72$$ 
aoVal: E
content: $$24$$ 
Answer: A
Analysis: The first student has six choices of books; the second has five; and the third has four. By the Rule of product, there is a total of $$6\times5\times4=120$$ways. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Use a simple method to compute the following.  $$25\times38$$=． 
Options:
aoVal: A
content: $$960$$ 
aoVal: B
content: $$950$$ 
aoVal: C
content: $$940$$ 
aoVal: D
content: $$930$$ 
Answer: B
Analysis: 25x38 =25x(30+8) =25x30+25x8 =750+200 =950 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Problem: Moon and Archie played chess competitively. Both of them had same levels of skill. They agreed to play seven games, and the one that win four games first would be the ultimate winner. Now, they have already played three games, and Moon won two games while Archie won one game. What is the probability that Mon be the ultimately winner? 
Options:
aoVal: A
content: $$\frac{3}{8}$$ 
aoVal: B
content: $$\frac{11}{16}$$ 
aoVal: C
content: $$\frac{3}{16}$$ 
aoVal: D
content: $$\frac{7}{16}$$ 
Answer: B
Analysis: $$\rm B$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: One day, Pip asks his parents: "What day is it today?"  His mother says: "Today is Monday."  His father says: "Today is Tuesday."  From the options below, \textbf{Pip agrees with B}. Do you agree with Pip? 
Options:
aoVal: A
content: One of these two sentences is definitely wrong and the other one is correct. 
aoVal: B
content: It is possible that both of Pip\textquotesingle s parents are wrong. 
aoVal: C
content: It is possible that both of Pip\textquotesingle s parents are right. 
aoVal: D
content: If Pip\textquotesingle s mother is wrong, then his father must be right. 
Answer: B
Analysis: "Today is Monday" is not the opposite of "Today is Tuesday".i.e. they can both be false.  "Today is Monday" is the direct opposite of "Today is not Monday". One must be true and the other must be false. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: Abe holds $1$ green and $1$ red jelly bean in his hand. Bob holds $1$ green, $1$ yellow, and $2$ red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (2013 AMC 8 Problem, Question \#14) 
Options:
aoVal: A
content: $\dfrac{1}{4}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{3}{8}$ 
aoVal: D
content: $\dfrac{1}{2}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: The probability that both show a green bean is $\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}$. The probability that both show a red bean is $\dfrac{1}{2}\times \dfrac{2}{4}=\dfrac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Two tiles numbered $1$ and $2$ are turned face down, respectively. One tile is turned up at random, and throw a die to get a number from $1$ to $6$. What is the probability that the product of the numbers on the tile and the die is greater than or equal to $12$? 
Options:
aoVal: A
content: $\frac {1}{2}$ 
aoVal: B
content: $\frac {1}{6}$ 
aoVal: C
content: $\frac {1}{3}$ 
aoVal: D
content: $\frac {1}{4}$ 
aoVal: E
content: $\frac {1}{12}$ 
Answer: E
Analysis: There are $12$ different combinations. The product of two numbers is greater than $12$ will be $2\times6$. Thus, the probability is $\frac 1{12}$ . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: What is the value of the following sum?  $$299 + 297 + 295 + 296 + 298$$ 
Options:
aoVal: A
content: 1494 
aoVal: B
content: 1490 
aoVal: C
content: 1485 
aoVal: D
content: 1484 
aoVal: E
content: None of the above 
Answer: C
Analysis: $$299 + 297 + 295 + 296 + 298$$ = $300 - 1 + 300 - 3 + 300 - 5 + 300 - 4 + 500 - 2$~ = $1485$~ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: There are $$29$$ students in a certain class. $$12$$ of the students have a sister and $$18$$ of the students have a brother. In this class, only Tania, Barbara, and Anna do not have any siblings. How many students from this class have both a brother and a sister? 
Options:
aoVal: A
content: $$$$None 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: $$6$$ 
Answer: D
Analysis: $(12+18)-(29-3)=4$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Eddie finished reading a story book last week. He read an average of $19$ pages per day for the first six days of the week and $26$ pages on the last day. How many pages on average did Eddie read per day? 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$20$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$6$$ 
Answer: B
Analysis: $(19\times6+26)\div7=140\div7=20$ pages. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: Julia, Kasia, Zuzanna, and Helena have their birthdays on March $$1^{\rm st}$$, May $$17^{\rm th}$$, July $$20^{\rm th}$$, and March $$20^{\rm th}$$. Kasia and Zuzanna were born in the same month. Julia and Zuzanna were born on the same day of a month. Which of the girls was born on May $$17^{\rm th}$$? 
Options:
aoVal: A
content: Julia 
aoVal: B
content: Kasia 
aoVal: C
content: Zuzanna 
aoVal: D
content: Helena 
Answer: D
Analysis: Given that Kasia and Zuzanna were born in the same month, their birth month must be March. Given that Julia and Zuzanna were born on the same day of a month, they must be born on the $$20^{\rm th}$$. Hence, Zuzanna was born on March $$20$$; Kasia was born on March $$1$$; and Julia was born on July $$20$$. Helena is therefore the one born on May $$17$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A cat divides $$24$$ fish into $$4$$ groups, and each group has at least $$1$$ fish. There are at most~\uline{~~~~~~~~~~}~fish in the group that has most of the fish. 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$1$$ 
aoVal: D
content: $$21$$ 
Answer: D
Analysis: $$ 1 + 1+1 + 21 = 24$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: What is the smallest possible sum of two positive integers with a product of $$100$$? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$52$$ 
aoVal: C
content: $$29$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: $$20$$ 
Answer: E
Analysis: Since the product of the two positive integers is $$100$$, the possible pairs of integers are $$\left( 1,100 \right)$$, $$\left( 2,50 \right)$$, $$\left( 4,25 \right)$$, $$\left( 5,20 \right)$$, $$\left( 10,10 \right)$$, the smaller the difference, the smaller the sum, so the smallest sum is $$10+10=20$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The diagram shows some cubes of the same size stacked at a corner of a room. How many cubes are there altogether? (Note: The floor is horizontal and the two walls are vertical. There are no gaps or holes behind the visible cubes). 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$25$$ 
aoVal: C
content: $$26$$ 
aoVal: D
content: $$27$$ 
aoVal: E
content: $$28$$ 
Answer: B
Analysis: Count layer by layer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: In how many ways can the letters in $CPCCKBY$ be rearranged so that $C$ cannot be put in both ends and two or more $C$s do not appear together? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$64$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$14$$ 
Answer: A
Analysis: There are $3$ $C$s in total now with other $4$ letters remaining.  There are $\_4P\_4$ ways for us to arrange the $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4=24$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: A drawer contains ten identical yellow socks, eight identical blue socks and one hundred identical pink socks. Amrita picks socks from the drawer without looking.  What is the smallest number of socks she must pick to be sure that she has at least two pairs of matching socks? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$13$$ 
Answer: B
Analysis: Consider the worst case scenario: first 3 are drawn, all of different colors, when the fourth is drawn, no matter which color is drawn, a pair of the same color will definitely be matched. Continue to draw the fifth one. The worst case is that the last color socks are drawn again, and at this point there are three socks of the same color and one sock of each of the other two colors. If you continue to draw the sixth one, no matter which color you draw, you will be able to form two pairs of socks of the same color. So the answer is to draw 6 times. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems
Problem: In a box, there are $10$ balls of which $4$ are red, $3$ are black and $3$ are white. What is the probability of picking up $3$ balls randomly such that there are $2$ red balls and $1$ balck ball? 
Options:
aoVal: A
content: $\frac12$ 
aoVal: B
content: $\frac13$ 
aoVal: C
content: $\frac1{12}$ 
aoVal: D
content: $\dfrac{3}{20}$ 
Answer: D
Analysis: $\dfrac{\_4C\_2 \times~~\_3C\_1}{\_{10}C\_3}=\dfrac{6\times3}{120}=\dfrac{3}{20}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: How many triangles of all sizes can be seen in the picture below?  [insert pic] 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$11$$ 
aoVal: D
content: $$13$$ 
aoVal: E
content: $$12$$ 
Answer: B
Analysis: [insert pic] 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Regular Figures->Counting Triangles
Source: 其它
Problem: Let $(a, b, c, d)$ be integers where they are not necessarily different. If each one of them is in the set ($0$, $1$, $2$, $3$), what is the probability that $a \cdot d-b \cdot c$ is odd? (For example, ($0$, $3$, $1$, $1$) meet the condition, because $0 \cdot 1-3 \cdot 1=-3$ is odd.) 
Options:
aoVal: A
content: $\frac3{16}$ 
aoVal: B
content: $\frac14$ 
aoVal: C
content: $\frac38$ 
aoVal: D
content: $\frac12$ 
aoVal: E
content: $\frac34$ 
Answer: C
Analysis: In order for $a \cdot d-b \cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\times2+2\times4=12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \cdot d-b \cdot c$ odd is $2 \cdot(12 \cdot 4)=96.$ Thus, the answer is $\frac{96}{256}=\frac38$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A positive integer divisor of $12 !$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$23$$ 
Answer: E
Analysis: The prime factorization of $12 !$ is $2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11$. This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12 !$. In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization.  Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in $12 !$. Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can choose $0,2,4,6,8,$ or $10$, etc.  The probability that the divisor chosen is a perfect square is $$ \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23 $$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: A pair of fair 6 -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals 7 at least once is greater than $\frac{1}{2}$ ? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
aoVal: E
content: $$6$$ 
Answer: C
Analysis: Rolling a pair of fair 6 -sided dice, the probability of getting a sum of 7 is $\frac{1}{6}$ : Regardless what the first die shows, the second die has exactly one outcome to make the sum 7 . We consider the complement: The probability of not getting a sum of 7 is $1-\frac{1}{6}=\frac{5}{6}$. Rolling the pair of dice $n$ times, the probability of getting a sum of 7 at least once is $1-\left(\frac{5}{6}\right)^{}n$. Therefore, we have $1-\left(\frac{5}{6}\right)^{}n\textgreater\frac{1}{2}$, or $$ \left(\frac{5}{6}\right)^{}n\textless\frac{1}{2} $$ Since $\left(\frac{5}{6}\right)^{4}\textless\frac{1}{2}\textless\left(\frac{5}{6}\right)^{3}$, the least integer $n$ satisfying the inequality is (C) 4 . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: There are $$20$$ red balls, $$2$$ black balls, and $$1$$ white ball in a bag. They are of identical shape, size and quality except for colour. Take out $$1$$ ball without looking. Among the following statements, which one is true? 
Options:
aoVal: A
content: The ball taken out must be a black ball. 
aoVal: B
content: It is impossible that a white ball will be taken out. 
aoVal: C
content: It is very likely that a red ball will be taken out. 
aoVal: D
content: It is certain that a red ball will be taken out. 
Answer: C
Analysis: The number of red balls in the bag is the most compared with other coloured balls. If we take out one ball at random, the probability of taking out a red ball is larger. So $$\text{C}$$ is the answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: There are some pieces of candy on a table.You are challenged by your friend to play the  following game: you both have to altemate moves, and in each move, you can take away either $$1$$, $$2$$, $$3$$,~ $$4$$ or $$5$$ pieces from the table. The person who takes away the final piece from the table wins. If you go second, how many pieces of candy should be on the table before the game starts such that you can ensure victory? 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$15$$ 
aoVal: C
content: $$18$$ 
aoVal: D
content: $$21$$ 
Answer: C
Analysis: Only $$18$$ is one of the multiples of $$5+1$$, and the second player can ensure victory. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Strategies and Operations->Game Strategy
Problem: Elsa is looking for some two-digit numbers. The difference between the digits in the tens place and the ones place is 5. How many of such numbers are there? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: 16, 27, 38, 49,  50, 61, 72, 83, 94 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: The faces of two identical fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? 
Options:
aoVal: A
content: $\dfrac{4}{9}$ 
aoVal: B
content: $\dfrac{1}{2}$ 
aoVal: C
content: $\dfrac{5}{9}$ 
aoVal: D
content: $\dfrac{3}{5}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$.  How to solve the problem: The probability of getting an odd number first is $\dfrac{4}{6}=\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\dfrac{2}{6}=\dfrac{1}{3}$. Now we multiply the two fractions: $\dfrac{2}{3}\times\dfrac{1}{3}=\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\dfrac{2}{9}\times2=\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\dfrac{4}{9}=\left (\text{C}\right )\dfrac{5}{9}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: There are many red balls, yellow balls, and blue balls of identical shape in a bag. Many students are playing a game in which each of them can take $$2$$ balls from the bag without observing the color of the balls. (After each student takes the balls, they should put them back in the bag.) The teacher finds that no matter how the students take the balls, there are always at least $$2$$ students who take the same balls. (For example, student $$A$$ and student $$B$$ both get one red ball and one yellow ball.) There would be at least~\uline{~~~~~~~~~~}~students playing the game. 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$7$$ 
aoVal: D
content: $$8$$ 
Answer: C
Analysis: There are $$6$$ possible cases when taking $$2$$ balls from the box. In the worst case, each of the first $$6$$ students takes different balls to the others. When the $${{7}^{\text{th}}}$$ student takes the balls, he/she would definitely take the same balls with someone among the first $$6$$ students. $$6+1=7$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: A city has a bicycle hire scheme where it is possible to hire a bicycle for short journeys.  Last year I hired a bicycle $$60$$ times and rode for $$13$$ hours altogether.  For how long on average did I hire the bicycle on each ride? 
Options:
aoVal: A
content: $$13$$ minutes 
aoVal: B
content: $$23$$ minutes 
aoVal: C
content: $$39$$ minutes 
aoVal: D
content: $$47$$ minutes 
aoVal: E
content: $$73$$ minutes 
Answer: A
Analysis: The rider travels for $$13$$ hours over $$60$$ rides, which is anaverage time of $$\frac{13}{60}$$ of an hour per ride, hence $$13$$ minutes. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Let $(a, b, c, d)$ be an ordered quadruple of not necessarily distinct integers, each one of them is in the set ($0$, $1$, $2$, $3$). What is the probability that $a \cdot d-b \cdot c$ is odd? (For example, ($0$, $3$, $1$, $1$) meet the condition, because $0 \cdot 1-3 \cdot 1=-3$ is odd.) (Adapted from 2020 AMC 10A Problem, Question \#18) 
Options:
aoVal: A
content: $\frac3{16}$ 
aoVal: B
content: $\frac14$ 
aoVal: C
content: $\frac38$ 
aoVal: D
content: $\frac12$ 
aoVal: E
content: $\frac34$ 
Answer: C
Analysis: In order for $a \cdot d-b \cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\times2+2\times4=12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \cdot d-b \cdot c$ odd is $2 \cdot(12 \cdot 4)=96.$ Thus, the answer is $\frac{96}{256}=\frac38$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: $$124$$ children attended a fair. $$87$$ of them tried out sports events. $$65$$ of them tried out art activities. How many children tried out both sport events and art activities? 
Options:
aoVal: A
content: $$22$$ 
aoVal: B
content: $$28$$ 
aoVal: C
content: $$37$$ 
aoVal: D
content: $$59$$ 
Answer: B
Analysis: Nil 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Professor Chang has nine language books lined up on a bookshelf: two different Arabic books, three different German books, and four different Spanish books. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together? ($2018$ AMC $8$ Problem, Question \#$16$)  $\textasciitilde$  $\textasciitilde$  $\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$  \hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}\hspace{0pt}$\textasciitilde$ 
Options:
aoVal: A
content: $$1440$$ 
aoVal: B
content: $$2880$$ 
aoVal: C
content: $$5760$$ 
aoVal: D
content: $$182440$$ 
aoVal: E
content: $$362880$$ 
Answer: C
Analysis: Since the two Arabic books and four Spanish books have to be kept together, respectively, we can treat them both as just one book. That means we\textquotesingle re trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $\_5P\_5$. Now we multiply this product by $\_2P\_2\times \_4P\_4$~because there are $\_2P\_2$~ways to arrange just two Arabic books, and $\_4P\_4$~ways to arrange just four Spanish books. Multiplying all these together, we have the answer $C$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它