Two state system probabilities

[anachin6000]

I started to study statistical physics from a book, and it starts with basics about statistics and probabilities (which are things mostly new for me).
In the book there is the following statement:

"The simplest non-trivial system which we can investigate using probability theory is one for which there are only two possible outcomes. There would obviously be little point in investigating a one outcome system. Let us suppose that there are two possible outcomes to an observation made on some system S. Let us denote these outcomes 1 and 2, and let their probabilities of occurrence be
P(1) = p,
P(2) = q.
It follows immediately from the normalization condition that
p + q = 1,
so q = 1 − p.
The probability of obtaining n1 occurrences of the outcome 1 in N observations is given by
PN(n1) = C_N (n1,N−n1) p^(n1) q^(N−n1), (2.16) where C_N (n1,N−n1) is the number of ways of arranging two distinct sets of n1 and N − n1 indistinguishable objects."

I'm familiar with combinatorics, so I find it obvious that their CN (n1,N−n1) is Cn1 N.
But I'm very curious how this can be proved. They give an example, but not a general demonstration. I've thought about it, but I couldn't do the demonstration. Can someone help me with it.
I must mention: it is not homework, it is just for my personal knowledge.

 

[PeroK]

I started to study statistical physics from a book, and it starts with basics about statistics and probabilities (which are things mostly new for me).
In the book there is the following statement:

"The simplest non-trivial system which we can investigate using probability theory is one for which there are only two possible outcomes. There would obviously be little point in investigating a one outcome system. Let us suppose that there are two possible outcomes to an observation made on some system S. Let us denote these outcomes 1 and 2, and let their probabilities of occurrence be
P(1) = p,
P(2) = q.
It follows immediately from the normalization condition that
p + q = 1,
so q = 1 − p.
The probability of obtaining n1 occurrences of the outcome 1 in N observations is given by
PN(n1) = C_N (n1,N−n1) p^(n1) q^(N−n1), (2.16) where C_N (n1,N−n1) is the number of ways of arranging two distinct sets of n1 and N − n1 indistinguishable objects."

I'm familiar with combinatorics, so I find it obvious that their CN (n1,N−n1) is Cn1 N.
But I'm very curious how this can be proved. They give an example, but not a general demonstration. I've thought about it, but I couldn't do the demonstration. Can someone help me with it.
I must mention: it is not homework, it is just for my personal knowledge.

You could prove it by induction. Have you come across mathematical induction?

 

[anachin6000]

You could prove it by induction. Have you come across mathematical induction?

Yes, I know about mathematical induction, but I was never good at using it. After your reply, I tried to it, but with no success.
First of all I tried to verify for 0, but it didn't verify, same for 1. I might miss something because I'm very tired.

 

[PeroK]

Yes, I know about mathematical induction, but I was never good at using it. After your reply, I tried to it, but with no success.
First of all I tried to verify for 0, but it didn't verify, same for 1. I might miss something because I'm very tired.

If you're tired, why not take a fresh look at it tomorrow.