In how many ways can a selection be done of $5$ letters? 
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$. 

$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $


Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way 
Number of ways to select $2$ similar and $3$ different letter = $\dbinom{4}{1}\times \dbinom{4}{3}=16$.
Number of ways of selecting $2$ similar + $2$ more similar letter and $1$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$. 
Number of ways to select $3$ similar and $2$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$. 
Number of ways to select $3$ similar and another $2$ other similar = $\dbinom{3}{1}\times \dbinom{3}{1}=9$ 
Number of ways to select $4$ similar and $1$ different letter = $\dbinom{2}{1}\times \dbinom{4}{1}=8$
Ways of selecting 
$5$ similar letters = $1$ 
Total ways = $1+16+18+18+9+8+1= 71$


Well I have the solution But I am not able to fully understand it.
Or if their could be an $\color{red}{\text{alternate way}}$ than it would be great.
I have studied maths up to $12$th grade.

Let $x_i$ be the number of letter $i$ chosen for $1\le i\le 5$, where we number the letters in alphabetical order.
We want to find the number of solutions in nonnegative integers to the equation $x_1+\cdots+x_5=5$
$\hspace{.5 in}$with the restrictions $x_1\le5,\; x_2\le4, \;x_3\le3,\; x_4\le2, \;x_5\le1$.
Let $S$ be the set of all solutions, and let $A_i$ be the set of solutions with $x_i\ge7-i$ for $i=2,\cdots,5$.
Using Inclusion-Exclusion, we have that
$\hspace{.15 in}\displaystyle\big|\overline{A_2}\cap\overline{A_3}\cap\overline{A_4}\cap\overline{A_5}\big|=|S|-\sum_{i}|A_i|+\sum_{i<j}|A_i\cap A_j|-\sum_{i<j<k}|A_i\cap A_j\cap A_k|+\cdots$
$\hspace{1.53 in}=|S|-|A_2|-|A_3|-|A_4|-|A_5|+|A_4\cap A_5|$
$\displaystyle\hspace{1.53 in}=\binom{9}{4}-\binom{4}{4}-\binom{5}{4}-\binom{6}{4}-\binom{7}{4}+\binom{4}{4}=\color{red}{71}$.
