An organization is assigned an IPv6 address block of 2001:db8:0:ca00::/56. How many subnets can be created without using bits in the interface ID space?

To determine how many subnets can be created from the given IPv6 address block of 2001:db8:0:ca00::/56 without using bits in the interface ID space, we first need to analyze the allocation of bits within the address structure.

An IPv6 address consists of 128 bits in total. In a /56 subnet, the first 56 bits of the address are designated for the network portion, leaving the remaining 72 bits for subnets and the interface ID. However, since we are tasked with creating subnets without using bits from the interface ID space, we focus solely on the bits available for subnets within the provided prefix.

In the case of a /56 prefix, the first 56 bits define the network, and we have the ability to manipulate the following bits for subnetting. Specifically, we have 8 bits left for creating subnets (from the 57th to the 64th bit).

To calculate the number of possible subnets, we take the number of bits available for the subnetting, which is 8 bits. Using the formula 2^n (where n is the number of bits allocated for subnetting), we find:

2^8 = 256.