[b]p1.[/b] Sujay sees a shooting star go across the night sky, and took a picture of it. The shooting star consists of a star body, which is bounded by four quarter-circle arcs, and a triangular tail. Suppose $AB = 2$, $AC = 4$. Let the area of the shooting star be $X$. If $6X = a-b\pi$ for positive integers $a, b$, find $a + b$.
[img]https://cdn.artofproblemsolving.com/attachments/0/f/f9c9ff23416565760df225c133330e795b9076.png[/img]

[b]p2.[/b] Assuming that each distinct arrangement of the letters in $DISCUSSIONS$ is equally likely to occur, what is the probability that a random arrangement of the letters in $DISCUSSIONS$ has all the $S$’s together?


[b]p3.[/b] Evaluate
$$\frac{(1 + 2022)(1 + 2022^2)(1 + 2022^4) ... (1 + 2022^{2^{2022}})}{1 + 2022 + 2022^2 + ... + 2022^{2^{2023}-1}} .$$


[b]p4.[/b] Dr. Kraines has $27$ unit cubes, each of which has one side painted red while the other five are white. If he assembles his cubes into one $3 \times 3 \times 3$ cube by placing each unit cube in a random orientation, what is the probability that the entire surface of the cube will be white, with no red faces visible? If the answer is $2^a3^b5^c$ for integers $a$, $b$, $c$, find $|a + b + c|$.


[b]p5.[/b] Let S be a subset of $\{1, 2, 3, ... , 1000, 1001\}$ such that no two elements of $S$ have a difference of $4$ or $7$. What is the largest number of elements $S$ can have?


[b]p6.[/b] George writes the number $1$. At each iteration, he removes the number $x$ written and instead writes either $4x+1$ or $8x+1$. He does this until $x > 1000$, after which the game ends. What is the minimum possible value of the last number George writes?


[b]p7.[/b] List all positive integer ordered pairs $(a, b)$ satisfying $a^4 + 4b^4 = 281 \cdot 61$.


[b]p8.[/b] Karthik the farmer is trying to protect his crops from a wildfire. Karthik’s land is a $5 \times 6$ rectangle divided into $30$ smaller square plots. The $5$ plots on the left edge contain fire, the $5$ plots on the right edge contain blueberry trees, and the other $5 \times 4$ plots of land contain banana bushes. Fire will repeatedly spread to all squares with bushes or trees that share a side with a square with fire. How many ways can Karthik replace $5$ of his $20$ plots of banana bushes with firebreaks so that fire will not consume any of his prized blueberry trees?


[b]p9.[/b] Find $a_0 \in R$ such that the sequence $\{a_n\}^{\infty}_{n=0}$ defined by $a_{n+1} = -3a_n + 2^n$ is strictly increasing.


[b]p10.[/b] Jonathan is playing with his life savings. He lines up a penny, nickel, dime, quarter, and half-dollar from left to right. At each step, Jonathan takes the leftmost coin at position $1$ and uniformly chooses a position $2 \le k \le 5$. He then moves the coin to position $k$, shifting all coins at positions $2$ through $k$ leftward. What is the expected number of steps it takes for the half-dollar to leave and subsequently return to position $5$?


PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. First, we need to determine the total number of distinct arrangements of the letters in "DISCUSSIONS". The word "DISCUSSIONS" has 11 letters where the letter 'S' appears 4 times, 'I' appears 2 times, and the other letters appear once each. The formula for the number of distinct permutations of a multiset is given by:
   \[
   \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
   \]
   where \( n \) is the total number of items to arrange, and \( n_1, n_2, \ldots, n_k \) are the frequencies of the distinct items. For "DISCUSSIONS", this becomes:
   \[
   \frac{11!}{4! \cdot 2!}
   \]
   Calculating the factorials, we get:
   \[
   11! = 39916800, \quad 4! = 24, \quad 2! = 2
   \]
   Therefore:
   \[
   \frac{11!}{4! \cdot 2!} = \frac{39916800}{24 \cdot 2} = \frac{39916800}{48} = 831600
   \]

2. Next, we need to determine the number of arrangements where all the 'S's are together. We can treat the block of 4 'S's as a single entity. This reduces the problem to arranging 8 entities: the block of 'S's, 'D', 'I', 'C', 'U', 'S', 'I', 'O', 'N'. The number of distinct arrangements of these 8 entities, considering that 'I' appears twice, is:
   \[
   \frac{8!}{2!}
   \]
   Calculating the factorials, we get:
   \[
   8! = 40320, \quad 2! = 2
   \]
   Therefore:
   \[
   \frac{8!}{2!} = \frac{40320}{2} = 20160
   \]

3. The probability that a random arrangement of the letters in "DISCUSSIONS" has all the 'S's together is the ratio of the number of favorable arrangements to the total number of arrangements:
   \[
   \frac{20160}{831600}
   \]
   Simplifying this fraction, we get:
   \[
   \frac{20160}{831600} = \frac{20160 \div 20160}{831600 \div 20160} = \frac{1}{41.25} = \frac{4}{165}
   \]

The final answer is \(\boxed{\frac{4}{165}}\).