For a given quantum statistical mechanical system with Hilbert space $\mathcal H$, hamiltonian $\hat H$, and number operator $\hat N$, the grand canonical partition function is defined as follows:
\begin{align}
  Z(\beta, \mu) = \mathrm{tr}(e^{-\beta(\hat H - \mu \hat N)})
\end{align}
where the trace is being taken over $\mathcal H$.

The difference between the statistics lies in the fact that for systems of identical particles, such as systems of bosons, the Hilbert space is smaller than if the particles are non-identical, so the trace is taken over a different vector space.

Suppose, for example, that we have a system of non-identical, non-interacting bosons.  Suppose, for the sake of simplicity, that the single particle Hilbert space is two-dimensional and that the single-particle Hamiltonian has two distinct eigenvalues $E_1$ and $E_2$ with corresponding orthonormal eigenstates $|1\rangle$ and $|2\rangle$.  Let $|0\rangle$ denote the state with no particles in it, then the Hilbert space of the system is spanned by the following energy eigenstates:
\begin{align}
  \text{0-particles}:&\qquad|0\rangle\\
  \text{1-particle}:&\qquad |1\rangle, |2\rangle\\
  \text{2-particles}:&\qquad |1\rangle|1\rangle\\ 
  &\qquad|1\rangle|2\rangle\\ 
  &\qquad|2\rangle|1\rangle\\ 
  &\qquad|2\rangle|2\rangle\\
  \text{3-particles}:
  &\qquad |1\rangle|1\rangle|1\rangle\\
  &\qquad |1\rangle|1\rangle|2\rangle\\
  &\qquad |1\rangle|2\rangle|1\rangle\\
  &\qquad |1\rangle|2\rangle|2\rangle\\
  &\qquad |2\rangle|1\rangle|1\rangle\\
  &\qquad |2\rangle|1\rangle|2\rangle\\
  &\qquad |2\rangle|2\rangle|1\rangle\\
  &\qquad |2\rangle|2\rangle|2\rangle\\
  &\qquad\vdots 
\end{align}
In fact, the $N$-particle subspace is spanned by $2^N$ distinct states.  However, if the particles are identical bosons, then the Hilbert space is spanned only by the symmetric combinations of states in each $N$-particle subspace, so listing the states gives:
\begin{align}
  \text{0-particles}:&\qquad|0\rangle\\
  \text{1-particle}:&\qquad |1\rangle, |2\rangle\\
  \text{2-particles}:&\qquad |1\rangle|1\rangle\\ 
  &\qquad \tfrac{1}{\sqrt{2}}\big(|1\rangle|2\rangle+ |2\rangle|1\rangle\big)\\
  &\qquad|2\rangle|2\rangle\\
  \text{3-particles}:
  &\qquad |1\rangle|1\rangle|1\rangle\\
  &\qquad \tfrac{1}{\sqrt{3}}\big(|1\rangle|1\rangle|2\rangle + |1\rangle|2\rangle|1\rangle + |2\rangle|1\rangle|1\rangle\big)\\
  &\qquad \tfrac{1}{\sqrt{3}}\big(|1\rangle|2\rangle|2\rangle + |2\rangle|1\rangle|2\rangle + |2\rangle|2\rangle|1\rangle\big)\\
  &\qquad |2\rangle|2\rangle|2\rangle\\
  &\qquad\vdots
\end{align}
and it is easy to show that in general, the subspace with $N$ identical bosons is spanned by $N+1$ distinct states.

In general, when the single-particle Hilbert space is greater than two dimensional, the counting may get more difficult, but the moral is the same.  You sum over less states when the particles are identical because the Hilbert space only includes appropriately symmetric or anti-symmetric states.