My answer comes rather late, so I don't expect up votes.
The response by user155957 was based on assuming that when you said 'combinations', you really meant 'permutations'. Based on this assumption, that user's response is correct by my reckoning.
Let us assume that when you said 'combinations', you really meant combinations.
In that case, the answer is 5.
If all the letters were distinct, such as they are in 'PENCIL', then the number of combinations possible with 6 distinct letters, taking 4 at a time, is given by the function nCr( 6, 4), which gives 15. However, the word is 'PEPPER', and the letters are not all distinct.
Writing the 'source partition' for the word PEPPER gives:
P E R
[3,2,1] <-- source partition
For combinations of 4 letters, we write all three possible 'target partitions' that go with [3,2,1]. These are:
[3,1]
[2,2]
[2,1,1]
Next to these, we list the corresponding selections (which are combinations, NOT permutations!)
[3,1] PPPE, PPPR
[2,2] PPEE
[2,1,1] PPER, EEPR
In conclusion, there are 5 combinations of 4 letters each that can be made from the word PEPPER.
A beginning:
There is no single formula for this. Therefore we have to organize the different types of cases, to which then known formulas apply.
There are $11$ slots where we have to write a letter. How many choices are there for the first two letters? They are of two different types, because of the double N. How many choices are there for the last two letters. Again there are different types.
Finally we have to fill in ("permute") the seven intermediate letters. There are no restrictions, but according to the types identified before the count will be different.
Best Answer
We could also use generating functions (which has recently been my favorite method since I suck at identifying cases).
Consider the multiset $M=[1\text{E}, 1\text{X}, 2\text{A}, 1\text{M}, 2\text{I}, 2\text{N}, 1\text{T}, 1\text{O}]$.
The number of $4$-permutations of $M$ is the coefficient of $x^4/4!$ in the exponential expansion of
$$(1+x)^5 \left(1+x+\dfrac{x^2}{2!}\right)^3.$$
The coefficient of $x^4$ above is $409/4$. Multiplying by $4!$ we get $2454$.
The $(1+x)$ represent the letters that are not repeated, while $(1+x+x^2/2)$ represents those letters that are repeated twice.