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.

Second line: You can get 2 of a type  from A,B,C or D so $4\choose 1$. Then ${4\choose3}$ to select 3 different letters from the 4 types not yet selected, and finally, apply multiplication principle.
Second last line: 4 of a type only from A or B, so ${2\choose1}$, 1 more selection from the remaining 4 types, ${4\choose 1}$, and multiply.
You should be able to understand the remaining along similar lines.
