Counting question / solution verification 
In a $100$ day period, each of six friends goes swimming on exactly $75$ days. There are $n$ days on which at least $5$ friends swim. Find the largest and smallest possible values of $n$.

My attempt at a solution:
The largest value of $n$ occurs when exactly $5$ friends swim on as many days as possible. 
We can achieve this by rotating the days on which each friend doesn't swim. If $A,B,C,D,E,F$ are the friends, then 


*

*$A$ doesn't swim on days $1,7,13,19,\ldots,85,91-100$ 

*$B$ doesn't swim on days $2,8,\ldots,86,91-100$ 

*$C$ doesn't swim on days $3,9,\ldots,87,91-100$ 

*$D$ doesn't swim on $4,10,\ldots,88,91-100$

*$E$ doesn't swim on $5,11,\ldots,89,91-100$

*$F$ doesn't swim on $6,12,\ldots,84,90,91-100$


Therefore, the largest possible value of $n$ is $90$.
The smallest possible value of $n$ occurs when exactly $4$ friends swim on as many days as possible. 
We can rotate these days as follows: $A$ doesn't swim on days $1,2$; $B$ doesn't swim on days $2,3$; $C$ doesn't swim on days $3,4$; $D$ doesn't swim on days $4,5$; $E$ doesn't swim on days $5,6$; $F$ doesn't swim on days $6,7$; $A$ doesn't swim on days $7,8;\ldots$ 
Continuing in this way, it is possible to ensure that exactly four people are swimming on days $1-75$ and all six people are swimming on days $76-100$. 
Therefore, the minimum value of $n$ is $25$.
Are these bounds correct? How could I improve my arguments?

There are $6 \cdot 75 = 450$ swimmer-days, so there is an obvious upper bound to the number of 5-swimmer days of $\frac{450}{5}=90$.  Your schedule is a concrete example of such, so 90 days must be the maximum.
A similar argument can be made about the lower bound.  There you want to maximize the number of days in which 2 or more friends are not swimming.  Any schedule with a day with 3 or more friends not swimming can be potentially improved by adjusting the schedule so that day has no more than 2 friends not swimming.  Any schedule with a day where exactly 5 friends are swimming can be potentially improved by having all 6 friends swim that day, so as to conserve non-swimming days.  So the best theoretical schedule uses has either 4 or 6 friends swimming each day, conserving the $6\cdot25=150$ nonswimming days to maximum effect to get 75 days with 4 or fewer friends swimming.  You also achieved this schedule.
