Find the number of ways you can invite $3$ of your friends on $5$ consecutive days. Find the number of ways you can invite $3$ of your friends on $5$ consecutive days, exactly one friend a day, such that no friend is invited on more than two days. 
My approach: Let $d_A,d_B$ and $d_C$ denote the total number of days $A, B$ and $C$ were invited respectively. According to the question we must have $0\le d_A,d_B,d_C\le 2.$ Also, we must have $$d_A+d_B+d_C=5.$$ 
Now let $d_A+c_A=2, d_B+c_B=2, d_C+c_C=2,$ for some $c_A, c_B, c_C\ge 0$. 
This implies that $c_A+c_B+c_C=1$. 
Therefore the problem translates to finding the number of non-negative integer solutions to the equation $$c_A+c_B+c_C=1.$$ 
By the stars and bars method the total number of required solutions is equal to $$\dbinom{1+3-1}{3-1}=3.$$
But the number of ways to invite the friends will be higher than this, since the friends are distinguishable and we have assumed them to be indistinguishable while applying the stars and bars method. 
How to proceed after this?

You can solve the problem as follows: 
Let's call the three friends as $A,B,C$, we need to invite them in such way that none of them go to your house more than 2 days. Obviously there are two friends (lest say $A$ and $B$) who will be invited two times and the other one only once. This way we can see that the problem consist in distribute $A,B,C$ in those five days. 
Choose $A$ first. You can invite him two days of the five days, so you can invite him in $\binom{5}{3}=10$ different ways. For $B$ you can invite him in $\binom{3}{2}=3$ because we can't invite him to go in those days that $A$ goes to your house. And for $C$ you have only one way to invite him (the day that is $A$ and $B$ are not coming to your house). Therefore, the are $10\times3\times1=30$ ways to invite your friends to your house, such that no friend is invited on more than two days.
