Why should we subtract 1 to get maximum number in bits

Question

I'm reading this article and it says that:

This means that an unsigned INT can go up to $4,294,967,296$ (which is $2^{32}$ $– 1$). You need to subtract one because the result of $2^32$ starts from $1$, while the first binary representation is $0$.

I have hard time understanding what is meant by the last sentence, can someone please explain?

Stick to 2 bits: the possibilities are 00,01,10,11. The number 4 would need an extra bit: 100. — almagest, May 04 '16 at 07:34
The biggest decimal value represented in four digits is $$9999 = 10^4 - 1$$ HTH. — CiaPan, May 04 '16 at 07:35
@almagest, so it's four combinations and it seems to be calculated using the formula of Permutations with Repetition mentioned here, correct? — Max Koretskyi, May 04 '16 at 07:46
Your confusion is in place. The statement "$4,294,967,296$ (which is $2^{32}–1$)" is plain wrong. — barak manos, May 04 '16 at 08:18
@barakmanos, that's what an article states. Do you care to elaborate? — Max Koretskyi, May 04 '16 at 08:30
That's why I said 'your confusion is in place'. I cannot elaborate on "$4,294,967,296=2^{32}-1$" in any other manner I'm afraid. — barak manos, May 04 '16 at 08:34
@barakmanos, ok, so the correct result ends in 295, not 296. Do you think that adds to my confusion? :) — Max Koretskyi, May 04 '16 at 17:20

score 3 · Answer 1 · answered May 04 '16 at 07:34

When you count up to $2^{32}$, you start counting $1$, $2$, ... then the $2^{32}$th number is $2^{32}$. Since the computer has to store the number $0$ in an unsigned int, it is actually starting to count with $0$, then $1$ and so on. That means that the $n$th number for the computer is, in fact, $n-1$. Hence, the biggest number it can store, the $2^{32}$th number is $2^{32}-1$

Why should we subtract 1 to get maximum number in bits

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