For each bit binary digit that you have, there are two possibilities: Either it can be a zero, or it can be a one. Therefore, if you have one bit, you have two possible numbers. It's based on this fundamental principle: So for the 32 bits mentioned, there are 2 choices for each bit.
For each choice of the first bit, there are 2 for the second. For each choice of the first and second, there are 2 for the third. For each choice of the first three bits, there are 2 for the fourth. You can think of it as a tree, with each row of the tree except the root corresponding to each bit position, starting from the rightmost though you could also start from the leftmost and going left.
There are two branches off each tree node corresponding to a choice of 1 or 0 for the given bit. How many nodes are there at the end of the tree? The next section is a proof from set theory, adapted from one I found in the cited document. If this is too advanced for you, then you can ignore it, but you kept making comments about wanting a "Formal Proof" in the other answers you got.
I'm not sure how else to satisfy this request. One can give a formal proof of the multiplication principle in a restated form from set theory.
The site gives a proof of this from the more fundamental principle. So for the multiplication principle. One can then use another simple inductive argument to extend to multiple sets. If you consider all zeroes as the starting point you then have one more addition - 4,,, becomes to total.
So from the all zeroes pattern counting on up to the all ones pattern gives four billion two hundred ninety-four million nine hundred sixty-seven thousand two hundred ninety-six different possiblities.
Join them; it only takes a minute: Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top. How many 32 digit binary number combinations are possible? I am not a mathematician. So, please try to explain in a simple way. Vishnu Vivek 4 11 People have shown how. What more do you want to make it "formal"?
By the fundamental counting principle see en. This easily extends to more than two choices. These are actual proofs from the principle just mentioned. You've re-written what I have explained in my solution in the question.
In this case proof by counting is logically defined. So, please try to explain in a simple way" There are 2 choices 0 or 1 for the first bit and 2 choices for the second bit. Epsilon 1, 1 7 Sign up or log in Sign up using Google.
This is a formal proof.More...