>In how many ways can 5 letters be selected from the word INSTITUTE? 
In how many ways can $5$ letters be selected and arranged from the word INSTITUTE?

MyApproach:
I N S T U E
$2$ $1$ $1$ $3$ $1$ $1$
From these I have to select $5$.I can have following cases.
Case1:
Taking $1$ from each and taking $1$ from either I  or T .
The given combination and arrangements can be $6C5$ . $5!$ 
Similarly,other cases can be 
Case 2: $1$,$1$,$1$,$2$ arrangements : $5!$/$2!$
Case 3: $1$,$1$,$3$   arrangements :$5!$/$3!$
Case 4: $1$,$2$,$2$   arrangements :$5!$/$2!$ . $2!$
Case 5: $2$,$3$     arrangements :$5!$/$2!$ . $3!$
I can calculate arrangements but I am unable to calculate selections for the Cases 2,3,4,5.

Can anyone give the hint about how to selection can be made in these parts? 

 A: Here $\bf{INSTITUTE}$ Contain $\bf{\left\{2I,N,S,3T,U,E\right\}}$
Now we will form Different cases
$\bullet\; $ If all letters are different, Then we will take $5$ different letters
So total no. of arrangement $\displaystyle = \underbrace{\binom{6}{5}}_{\bf{select\; 5\; diff.\; letters}}\times \underbrace{5!}_{\bf{arrange\; these\; 5\; letters}}$
$\bullet\; $ If $2$ letters are same and other $3$ letters are different.
So we will take a $1$ pair of letter from $2$ pair letter and other $3$ from $5$ different letters
So tatal no. of ways $\displaystyle =\underbrace{\binom{2}{1}\times \binom{5}{3}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{5!}{2!}}_{\bf{arrangement\; of\; letters}}$
$\bullet\; $ Here $2$ same letters are of same kind and other $2$ are of same kind
So tatal no. of ways $\displaystyle \underbrace{\binom{2}{2}\times \binom{4}{1}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{4!}{2!\times 2!}}_{\bf{arrangement\; of\; letters}}$
$\bullet\; $ Here $3$ same letter are of same kind and other $2$ letter are different
So tatal no. of ways $\displaystyle = \underbrace{\binom{1}{1}\times \binom{5}{2}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{5!}{3!}}_{\bf{arrangement\; of\; letters}}$
$\bullet\; $ Here $3$ same letter are of same kind and other $2$ letter are of same kind
So tatal no. of ways $\displaystyle = \underbrace{\binom{1}{1}\times \binom{1}{1}}_{\bf{selection \; of \; letters}}\times \underbrace{\frac{5!}{2!\times 3!}}_{\bf{arrangement\; of\; letters}}$
So Total no. of ways $ = $ add all above cases.
