simple question about cryptographic hash function and its preimage collision

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Hi all,
I am reading "cryptography and network security principles and practice", in chapter 11.3, the book says:
"Let us consider how many preimages are there for a given hash value, which is
a measure of the number of potential collisions for a given hash value. Suppose the length of the hash code is n bits, and the function H takes as input messages or data blocks of length b bits with . Then, the total number of possible messages is 2^b and the total number of possible hash values is 2^h. On average, each hash value corresponds to 2^(b/n) preimages. If H tends to uniformly distribute hash values then, in fact, each hash value will have close to 2^(b/n) preimages."

So my question is that isn't it obvious that each hash value corresponds to 2^(b-n) preimages since 2^b / 2^b = 2^(b-n), please correct me if I am wrong !

Thanks in advance !