Wow!  What a difficult problem!

There are 2 D's, 3 E's, 1 I, 3 N's, 1 P, and 1 T

I hope you really meant combinations, and not permutations.  For I am not
considering taking the letters in any order.  For example, 
"E,E,N,N,N", "E,N,E,N,N", "N,N,N,E,E", and "E,N,N,N,E"  
are all counted as the same combination, even though they would be different
permutations.  Let me know in the thank-you note if you meant permutations
instead, and I'll re-work the problem.

There are 5 basic cases of choosing 5 letters.  I will use the last five
letters of the alphabet, V,W,X,Y and Z as "placeholders" for the letters of a
combination.  Here are the 5 types of combinations:

1. V,V,W,W,W  1 pair, both of the same letters, and 1 triplet, all of the same letter. 
2. V,V,W,W,X  2 pairs of same letters and 1 non-matching letter.
3. V,V,W,X,Y  1 pair of the same letter and 3 non-matching letters. 
4. V,V,V,X,Y  1 triplet, all of the same letter and 2 non-matching letters.
5. V,W,X,Y,Z  5 non-matching letters.

Case 1.  V,V,W,W,W

W is more restrictive here than V so we choose it first.
We can choose the letter for W in 2 ways, either E or N
We can choose the letter for V in 2 ways, D or whichever one of {E,N} we didn't
choose in the preceding sentence.
That's 2x2 or 4 combinations for case 1.

Case 2.  V,V,W,W,X

We can choose the 2 letters for the 2 pairs from {D,E,N} in C(3,2) = 3 ways
We can choose the letter for X in any of the remaining 4 ways.
That's 3x4 or 12 combinations for case 2.

Case 3. V,V,W,X,Y

We can choose the letter for the pair 3 ways, D,E, or N
We can choose the letters W,X,Y from any of the 5 remaining ways, in C(5,3) or
10 ways
That 3x10 or 30 combinations for case 3.

Case 4. V,V,V,X,Y

We can choose the letter for V in 2 ways, E or N.
We can choose X,Y from any of the remaining 5 letters, in C(5,3) or 10 ways.
That's 2x10 or 20 combinations for case 4.

Case 5. V,W,X,Y,Z
That's 6 letters taken 5 at a time, or C(6,5) = 6 ways

Answer: 4+12+30+20+6 = 72

Edwin