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.

If Order is Unimportant
The number of ways to choose $5$ letters (if their order is unimportant) is the coefficient of $x^5$ in
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{x^2}}^{1\text{ E}}
\overbrace{\left(1+x+x^2\right)}^{2\text{ D's}}
\overbrace{\left(1+x+x^2+x^3\right)}^{3\text{ C's}}
\overbrace{\left(1+x+x^2+x^3+x^4\right)}^{4\text{ B's}}
\overbrace{\left(1+x+x^2+x^3+x^4+x^5\right)}^{5\text{ A's}}\\
&=\frac{1-x^2}{1-x}\frac{1-x^3}{1-x}\frac{1-x^4}{1-x}\frac{1-x^5}{1-x}\frac{1-x^6}{1-x}\\
&=\frac{1-x^2-x^3-x^4+O\left(x^7\right)}{(1-x)^5}\\[3pt]
&=\small\left[1-x^2-x^3-x^4+O\!\left(x^7\right)\right]\!\left[1+5x+15x^2+35x^3+70x^4+126x^5+210x^6+O\!\left(x^7\right)\right]\\[9pt]
&=1+5x+14x^2+29x^3+49x^4+71x^5+90x^6+O\!\left(x^7\right)\tag{1}
\end{align}
$$
where we used the Binomial Theorem for $(1-x)^{-5}$ above.

The coefficient of $x^5$ in $(1)$ is $71$.


If Order is Important
If the order of the letters is important, we can compute the exponential generating function with
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{\frac{x^2}{2!}}}^{1\text{ E}}
\overbrace{\left(1{+}x{+}\frac{x^2}{2!}\right)}^{2\text{ D's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}\right)}^{3\text{ C's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3}{+}\frac{x^4}{4!}\right)}^{4\text{ B's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}{+}\frac{x^4}{4!}{+}\frac{x^5}{5!}\right)}^{5\text{ A's}}\\[3pt]
&=\small1+5x+24\frac{x^2}{2!}+111\frac{x^3}{3!}+494\frac{x^4}{4!}+2111\frac{x^5}{5!}+8634\frac{x^6}{6!}+O\left(x^7\right)\tag{2}
\end{align}
$$

The coefficient of $\frac{x^5}{5!}$ in $(2)$ is $2111$.

