Recurence relation

Question

Generally a recurrence relation is given as

T(n)=aT(n/b)+ f(n)

where a is no of sub prblm of size n/b each.

Can there be any real scenario where a>b.

I mean if n is divided into n/b problems then there must be at most b problems...So that b*n/b=n.....then how can a>b??

Comments

This is the representation of a recursive eqn which is dividing a problem of size n into a number of subproblems of size b.

Now it's not a necessary condition that a should be greater than b.
for ex:

In merge sorting we sort n number recursively by dividing them half thus it comes out to be 2T(n/2)+n

where as

In binary search we reduce problem recursively Like : T(n/2) + 1...