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?
 A: A hint:
The only configurations that obey your constraint are:  
person A:  2 days
person B:  2 days
person C:  1 day
(We'll assign names to these different people below.)
Suppose you start with the "1 day" person.  Then there are just two legal sequences:
CABAB and CBABA
Suppose you start instead with a "2 day" person (e.g., A).  Write out the sequences to see there are just $6$ legal sequences.
ABABC, ABACB, ... 
But you could interchange the names of these people:  
Mary = A, Tom = B, Chris = C.  
OR 
Tom = A, Mary = B, Chris = C 
OR 
....
Check these combinations and add up!
Hope that helps.
A: The fact that the only way to achieve this is by inviting one friend over once and the other two twice will make this problem simpler.
How many ways are there to pick the one friend (from $3$) that will be only visiting one day instead of two?
Now if we assume the days are Monday through Friday, how many ways are there to pick the day that the one-day friend will visit?
Finally, for the remaining four days, how many ways are there to pick two of them? This will be the number of ways to arrange the remaining two friends.
Now multiply the three of these numbers together for the final answer 

A side proof as to why we must have it with one friend one day and the other two with $2$: We obviously can it have any friend arrive three days out of the five, as per given in the question. If all three come twice, this would require $6$ days so this isn't possible either. If two friends only come one day each, then we aren't using up all $5$ days.
A: 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.
