$1$. Five students each take a mock exam, following which the exam scripts are redistributed among the students so that each student marks one script. How many possible ways are there to redistribute the scripts so that no student marks his or her own script?
I applied the inclusion-exclusion principle. I got $5! - 4!5C1 +3!5C2 -2!5C3 + 1!5C4 -0!5C5 = 44$. Is this a correct application of the inclusion-exclusion principle ? Could I do this instead by considering the cycle type of presents maybe ?
$2$. I have a $10$-sided die whose faces are labeled with the numbers $1$ to $10$. When rolled, each outcome is equally likely to occur. I roll the die $5$ times in succession. Give, with justification, exact expressions for:
(i) The probability that I roll at least one $10$.
(ii) The probability that two of the rolls are each at most $3$ and the other three rolls are each at least $6$.
(iii) The probability that each roll is strictly greater than the previous roll.
(i) Total possibilities - $10^5$. Possibilities for no $10$s = $9^5$, so probability = $\frac{10^5-9^5}{10^5}$.
(ii) I am most unsure on this one. I considered $5$ separate sets for each roll. The first $2$ sets contain the numbers $1,2,3$, and the final $3$ sets contain the numbers $6,7,8,9,10$. The total number of possibilities hence are $3^2 \cdot 5^3$. So probability is $\frac{3^2 \cdot 5^3}{10^5}$.
(iii) I think this is just a combination with no repetition of a subset of size $5$, giving $10C5$, so probability is $\frac{10C5}{10^5}$.
Can someone check my solutions and let me know my errors ? : )
$44$ is correct, and your application of PIE is correct. You could also do this by considering cycle types. There are two possibilities for the cycle type of a derangement on $5$ symbols; either a $2$-cycle and a $3$-cycle, or a single $5$-cycle, as these are the only ways to have numbers which are $2$ or more summing to $5$. There are $5!/5=24$ possible $5$-cycles, and $\binom52 \cdot (2!/2)\cdot (3!/3)=20$ ways to have a two-cycle and a three-cycle.
Your work for $(i)$ and $(iii)$ is correct.
For $(ii)$, you are undercounting. You are only counting sequences like $(s,s,b,b,b)$, where $s\in \{1,2,3\}$ and $b\in \{6,7,8,9,10\}$. However, you need to also account for the other possible orderings, like $(b,s,b,s,b),(b,s,b,b,s)$, etc. You can fix your answer by multiplying by the number of possible ways that the two $s$'s and three $b$'s can be arranged in a row.