Number of partitions of 11 elements into 3 subsets?

If you substitute into that formula correctly then you get $\begin{Bmatrix} 4 \\ 3 \end{Bmatrix} = \dfrac{1}{6} \bigg( (1)(1)(3^4) + (-1)(3)(2^4) + (1)(3)(1^4) + (-1)(1)(0^4) \bigg)$, which correctly gives 6; I'm not sure why you stopped at the first term of the sum.

Sorry to bother you again I just want to make sure I've fully understood! I've just tried doing it with n=11, k=3;

$\frac{1}{6}[((1)(1)(3^{11})) + ((-1)(3)(2^{11})) + ((1)(3)(1^{11})) + ((-1)(1)(0^{11}))]$

However, being to the power of 11, I feel like this sequence should carry on? But $\displaystyle \binom{3}{4}$ and $\displaystyle \binom{3}{5}$ etc are impossible so I might be right in ending at $\displaystyle \binom{3}{3}$ in the final $(-1)$$\displaystyle \binom{3}{3}$$(0^{11})$ = $(-1)(1)(0^{11})$ term?

Thank you again!

Why would the series carry on? The series runs from 0 to k, and in this case k=3, so it makes sense that you have 4 terms. The fact that you have a power of 11 in there needn't have an impact on the sum (and if the 11 wasn't there then it would be worrying because it doesn't appear anywhere else!) From what I can see, you've done it right; you just have to evaluate it now.