$a+b+c+d+e+f=14$ , where $a,b,c,d,e$ and $f$ are whole numbers $\le 4$

This problem is a perfect demonstration of stars and bars. We can think about distributing the units of 14 into 6 baskets, each representing one of the variables.

So, we divide our pile of 14 units (*) into 6 pieces (|). This leaves us with

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So, we choose where to place the 5 bars in the 19 possible positions, so the answer is $19\choose{5}$ which is $11628$.

Now, I've ignored the condition that none of the numbers can be greater than 4. In order to account for that, I'll count the number of ways that one number will be greater than 4. First, I have 6 choices as to where to put 5 of our 14 units, and then, the remaining combinations are

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So, we have $6\cdot{14\choose5}$. However, we've now undercounted, since if two of the numbers are above 4, then we double subtracted them from the total.

Hence, assume that we choose two numbers $6\choose2$ to have value 5 or above. So, we distribute these values and spread the remaining 4 across the bars

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This gives us $15{9\choose4}$.

So, our answer is ${19\choose 5} - 6{14\choose5} + {6\choose2}{9\choose4} = 1506$.

Another method is using generating functions (e.g. here and here) leading to $$(1+x+x^2+x^3+x^4)^6$$ with the coefficient of $\color{red}{x^{14}}$ being the answer, which is 1506.