Where am I mistaken in my approach through PIE? 
A man has 3 friends. The number of ways in which he can invite one
  friend everyday for dinner on six successive nights so that no friend
  is invited more than three times is?

I am trying to use Principle of Inclusion and Exclusion to solve this problem. 
Attempt: 
$N(\bar{c_1}\bar{c_2}\bar{c_3})= N - (N(c_1)+N(c_2)+N(c_3)) + (N(c_1c_2)+N(c_1c_3)+N(c_2c_3)) - (N(c_1c_2c_3))$
$c_i$= condition that i$^{th}$ friend is invited more than thrice. 
$c_ic_j$ = $i^{th}$ AND $j^{th}$ friend are invited more than thrice...
$\implies \left(\dbinom{3}{1}\right)^6- 3\times(\dbinom{6}{4}\dbinom{2}{2}\times 2! + \dbinom{6}{5}\dbinom{1}{1}\times 2!+1)+0-0= 630$
Because:$N(c_1)= N(c_2)= N(c_3)$ and, explanation of the rest of the part (the part in brackets that is multiplied by $3$): 
There are only three possibilities, one of the friends is invited either for $5$ days or for $6$ days or for $4$ days. 
So I chose any $4$ days out of $6$ days, gave them to the friend who is being invited extra times and then $2$ days out of remaining two days and arranged them between two friends. 
then, I chose $5$ days out of $6$ days, gave them to the friend who is getting extra dinners and  arranged the remaining one day among the remaining two friends. 
Lastly, when a friend is invited for all 6 days, the remaining friends have $0$ choices. 
But answer given is $510$. Where have I gone wrong? 
 A: Since the man only invites one friend to dinner each night, it is not possible for him to invite two different friends more than three times each since $2 \cdot 4 = 8 > 6$.  Hence, the events we need to exclude are inviting the same friend to dinner on four, five, or six days.
If there were no such restriction, the man would have three options on each of the six days, so there would be $3^6$ possible dinner schedules. 
The same friend is invited to dinner exactly four times:  There are $3$ ways to choose the friend that is invited to dinner four times, $\binom{6}{4}$ ways to choose the days that friend is invited, and $2^2$ ways to invite one of the other two friends to dinner on the remaining two days.  Hence, there are 
$$\binom{3}{1}\binom{6}{4}2^2$$
such dinner schedules.
The same friend is invited to dinner exactly five times:  There are $3$ ways to choose the friend that is invited to dinner five times, $\binom{6}{5}$ ways to choose the days that friend is invited, and $2$ ways to choose which of the other friends is invited on the other day.  Hence, there are 
$$\binom{3}{1}\binom{6}{5}2$$
such dinner schedules.
The same friend is invited to dinner on all six days:  There are three ways to choose the friend who is invited to dinner every day.
Hence, the number of ways the man can invite one of his three friends to dinner each day for six days without inviting the same friend to dinner on more than three days is 
$$3^6 - \binom{3}{1}\binom{6}{4}2^2 - \binom{3}{1}\binom{6}{5}2 - \binom{3}{1}$$
A: Lets call the friends $a,b,c$ and multinomially expand
\begin{eqnarray*}
(a+b+c)^6 = \color{blue}{\sum a^6+6 \sum a^5 b +15 \sum a^4 b^2 + 30 \sum a^4 bc} \\+ \color{red}{20 \sum a^3 b^3 + 60 \sum a^3 b^2 c +90 a^2 b^2 c^2}.
\end{eqnarray*}
The terms in blue corespond to the events we are trying exclude and the terms in red correspond to the events we are trying to count.
Now lets count them 
\begin{eqnarray*}
\underbrace{(a+b+c)^6}_{\color{blue}{729}} = \underbrace{\sum a^6}_{\color{blue}{3}}
+\underbrace{6 \sum a^5 b}_{\color{blue}{36}} +\underbrace{15 \sum a^4 b^2}_{\color{blue}{90}} + \underbrace{30 \sum a^4 bc}_{\color{blue}{90}} \\+ \underbrace{20 \sum a^3 b^3}_{\color{red}{60}} + \underbrace{60 \sum a^3 b^2 c}_{\color{red}{360}} +\underbrace{90 a^2 b^2 c^2}_{\color{red}{90}}.
\end{eqnarray*}
So inclusion exclusion would give $\color{blue}{729-3-36-90-90}=510$, alternatively we can just count the events we want $\color{red}{60+360+90}=510$.
