<s>[INST] 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.
 [/INST] Ok so we are breaking this down by case:
Case 1: none of our letters are the same and we pick 5 distinct letters

Case 2a: exactly 2 of the letters are the same and the other 3 are distinct.

Case 2b: 2 of the letters are the same, 2 others are the same, and the fifth is distinct.

Case 3a: exactly 3 of the letters are the same and the other 2 are distinct.

Case 3b: exactly 3 of the letters are the same and the other 2 are also the same.

Case 4: exactly 4 of the letters are the same and the other 1 is distinct.

Case 5: all 5 letters are the same.

We see that any combination must be one of these $5$ cases, and there is no overlap in the $5$ cases. Thus we are going to sum up the number of total possibilities for each case.

Case 1: If we pick $5$ distinct letters, we pick one of each letter, so there is only $1$ way to do this.
Case 2a: First lets count the ways to pick the $2$ letters which are the same. It can't be E, so there are only $4$ ways to pick them. Second, we must pick $3$ more from the remaining $4$ letters (none of these can be the same letter as the first $2$ since that would be a different case). Thus we use $4$ choose $3$, which turns out to be $4$ options. We multiply these together to see that we have $4\times 4=16$ options for case 1.
Case 2b: We are now picking two pairs from four possible letters which have sufficient quantities. Thus the ways to pick the two pairs is $4$ choose $2$. The fifth letter is any of the remaining $3$ letters, so we have $6\times 3=18$ for this case.
Case 3a: Similarly, there are only $3$ ways to pick $3$ matching letters. Then we proceed to pick the other $2$ from $4$ choices, so we multiply $3$ by $4$ choose $2$ to get $3\times 6=18$ options for case 3.
Case 3b: There are $3$ ways to pick the triplet and the other pair only has $3$ options also, since it can't be the same letter as the triplet (that would be case 5). Thus there are $3\times 3=9$ possibilities.
Case 4: Hopefully you follow the pattern: there are only $2$ ways to pick $4$ of the same letter, and $4$ ways to pick the fifth letter, so we have $2\times 4=8$ for this case.
Case 5: Finally, there is only one way to pick $5$ of the same number.

Summing those together we get $1+16+18+18+9+8+1=71$ options. For me, this is the most straightforward approach once you understand it.
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