Number of ways so that exactly one permutation of the word TIDE occurs I want to calculate the number of $8$ letter words that can be formed using the letters of the word $TIDE$. However, in any word only one permutation of the word $TIDE$ should be present. That means among the $8$ letters I should not have four twins. 
For example, $ITDETTDD$ is a valid word since I have only one permutation of $TIDE$ and I cannot get other if i remove already obtained Permutation. But the word $TTIIEEDD$ or $TDEIDEIT$ is not valid since I can get two possible permutations.
My try:  we have $8$ positions. Each position can have $4$ choices from $T$, $I$, $D$ and $E$. 
So total words =$4^8$.
From this if we remove all the words of the form $TTIIEEDD$ we get all the words that contain only one string of $TIDE$.
Now all the words of the form $TTIIEEDD$ is $\frac{8!}{2!2!2!2!}$
So required answer is $$4^8-\frac{8!}{2!2!2!2!}=63016$$
EDIT: Well i got a different way to do it:
Since each letter must appear atleast once,In four places if we fix the letters $T$, $I$, $D$, $E$. then remaining four places have the following possible cases
Case $1.$ Words of the form $TIDETTTT$ which means eaxctly one letter repeated four times in other four places. So number of such words is
$$\binom{4}{1} \times \frac{8!}{5!}$$
Case $2.$ words of the form $TIDETTTD$ where exactly two letters occupy in other four places, but one repeats thrice.So all such words are
$$\binom{4}{2} \times 2 \times \frac{8!}{4!2!}$$
Case $3.$ Words of form $TIDETTDD$ where exactly two letters occupy in other four places,but each repeated twice. Number of such words are
$$\binom{4}{2} \times \frac{8!}{3!3!}$$
Case $4.$ Words of form $TIDETTDE$ where exactly three letters occupy other four places and among them one repeats twice. Number of such words are
$$\binom{4}{3} \times 3 \times \frac{8!}{3!2!2!}$$
Hence final answer will be sum of all the above  which is
$$8! \left(\frac{4}{120}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}\right)=38304$$
 A: The wording of the problem is unfortunate, since the usual meaning of string is sequence of consecutive letters. 
You have dealt with the "TIDE does not occur twice" correctly. However, you have not counted the words in which TIDE occurs at least once correctly, for there certainly are not $4^8$ such words.
To count them, we can either divide into cases or use Inclusion/Exclusion. We carry out most of the Inclusion/Exclusion process. There are $4^8$ words. Let us count the bad words, in which at least one of the letters is missing. There are $3^8$ words in which T is missing, and the same with the other letters. If we add up, getting $4\cdot 3^8$, we are double-counting the bad words in which, for example, T and I are both missing. There are $2^8$ such words, and there $\binom{4}{2}$ ways to choose the two missing letters.
So our next estimate of the number of bad words is $4\cdot 3^8-\binom{4}{2}2^8$. However, we have subtracted too  much, for we have subtracted one too many times the words in which for example all of T, I, D are missing. 
Now put things together. 
A: There are 65536 total combinations of the available letters and there are 2520 combinations containing two of each letter.  Therefore:
65536-2520 = 63016
EDIT#1:
If we must also exclude those combinations that lack a letter (24712) then:
65536-2520-24712 = 38304
There is no overlap in the two sets of exclusions so they can be directly subtracted.
EDIT#2:
To generate the exclusion counts, I used an Excel spreadsheet:

I placed all 65536 combinations in columns A through H.  Column I represents the four twins exclusion:
In I1
=IF(AND(COUNTIF(A1:H1,"T")=2,COUNTIF(A1:H1,"I")=2,COUNTIF(A1:H1,"D")=2,COUNTIF(A1:H1,"E")=2),1,0)

and copy down.  Column J represents the missing letter exclusion; so in J1:
=IF(OR(COUNTIF(A1:H1,"T")=0,COUNTIF(A1:H1,"I")=0,COUNTIF(A1:H1,"D")=0,COUNTIF(A1:H1,"E")=0),1,0)

and copy down..........the exclusion values are just the simple sums of columns I and J.
A: This is tedious but.
Case 1: There is one T,I,D,E and 4 Es.  Or in other words one T,I,D and 5 Es.  There are 8*7*6 ways to place the T,I,D so there are 8*7*6 = 8!/5! total ways to do this.
Case 1b: The E was arbitrary.  4*8!/5! ways to have 5 of a single letter.
Case 2: There is one T, I, D, E.  3 E's an extra D.  Or in other words one T,I, 2 Ds and 4 E.  There are 8*7 = 8!/6! ways to place the T, I.  There are there ar ${6 \choose 2}$ ways to place the Ds.  So 8!/6!*6!/4!2! = 8!/4!2! total..
Case 2b: the D and E were arbitrary so $(4*3)*8!/4!2!$ ways to do 4 of a letter and 2 of another.
Case 3: TIDE plus 2 Ds and 2 Es.  Or 1 T,I and 3 D's and 3 Es.  Again 8!/6! ways to do the T,I.  And ${6 \choose 3} ways to do the D.  So  8!/3!5! total.
Case 3a: Arbitrary:  4*3 chooses for which two letters are tripled and 4*3/2 as order of choosing the 2 doesn't matter.  So (4*3/2)8!/3!5! total.
Case 4: TIDE plus and an extra IDE plus an extra E.  Or in other words two of every letter but T and 3 Es.  There's ${ 8 \choose 3}$ ways to place the Es.  ${5 \choose 2}$  ways to place the Ds, and ${3 \choose 2} = 3$ ways to place the Is.  So $8!/5!3! * 5!/3!2! * 3!/2! = 8!/3!2!2!$.
Case 4a: Arbitrary: 4 choose for the triple letter; 3 for the single letter.
$(4*3)8!/3!2!2!$ total.
So total is $ 8!/4!2! +  (4*3)*8!/4!2! + (4*3/2)8!/3!5! +  (4*3)8!/3!2!2!$
$= 8!(1/4!2! + 12(1/4!2! + 1/2*3!5! + 1/3!2!2!))$
$= 8*7*3*5 + 4*7*3*5 + 2*3*8*7 + 8!/2$
$= 28(60 + 15 + 12 + 6*5*4*3*2) = 28(87 + 720) = 22596$.  I think.
