Find the number of ways when $4A, 1B, 1C, 1D$ is distributed among $4$ students with a few conditions 
And one of the answers I saw was along the following lines: 
By Principle of Inclusion-Exclusion, it follows that $4^3({4\choose 3}-4)-4\cdot3^3({6\choose 2}-3)+6\cdot 2^3 ({5\choose 1}-2) =832$.
Can anyone explain me the process involved in the PIE?
 A: Without any restrictions there are $4^3\binom73$ possibilities
Give the persons the numbers $1,2,3,4$ and $E_i$ denote the distributions in which person $i$ gets all $4A$, and let $F_i$ denote the distributions that person $i$ gets no letter.
Then to be found is: $$|E_1^{\complement}\cap E_2^{\complement}\cap E_3^{\complement}\cap E_4^{\complement}\cap F_1^{\complement}\cap F_2^{\complement}\cap F_3^{\complement}\cap F_4^{\complement}|=$$$$4^3\binom73-|E_1\cup E_2\cup E_3\cup E_4\cup F_1\cup F_2\cup F_3\cup F_4|$$
We discern the following relevant cases:


*

*$4$ times $|E_1|=4^3$

*$4$ times $|F_1|=3^3\binom62$ 

*$12$ times $|E_1\cap F_2|=3^3$

*$6$ times $|F_1\cap F_2|=\binom512^3$

*$12$ times $|E_1\cap F_2\cap F_3|=2^3$

*$4$ times $|F_1\cap F_2\cap F_3|=1$

*$4$ times $|E_1\cap F_2\cap F_3\cap F_4|=1$


$$4^3\binom73-4\cdot\left[4^3+3^3\binom62\right]+12\cdot3^3+6\cdot\binom512^3-12\cdot2^3+4\cdot1-4\cdot1=$$$$2240-1876+324+240-96+4-4=832$$
So the same outcome and quite some recognizable factors and terms.
