Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$\text{HH}, \text{HT}, \text{TH}, \text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $\text{HH}$ and $\text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $\text{HT}$ and $\text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$\text{two heads}, \text{two tails}, \text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.