No ways to arrange ABC by using C(3,2) is AB BC AC . Not BA CB OR CA including.
When we solve $(a+b)^2$
a^2 + b^2 + 2ab
Is it that no ways to arrange for the 3rd term since it would be C(2,1) . it is no ways to arrange them would be ab or ba. How do we write this here then?
On
Yes, the third term, $2ab$ stems from the fact that the # of ways of secting one element from a choice of 2 is $\binom{2}{1} = \frac{2!}{1!\times(2-1)!} = 2.$
Write down $(a + b)$.
Directly below it, write down $(a + b)$ again.
When you manually multiply it out, you see that the $2ab$ term comes from :
First you select one of the $a$'s from either the first or second line.
Once that $a$ is chosen, considering that you are focusing on the $2ab$ term, you will then choose the $b$ from the other line.
For a clearer understanding of why combinations rather than permutations are involved, perform the same manual computation of $(a + b)^3$ by writing $(a + b)$ on three separate lines, and work through the multiplication.
For example, the $3a^2b$ term will arise because there are $\binom{3}{2}$ ways of choosing which 2 of the 3 lines, you will select the $a$ factor from. Then, each selection of 2 $a$'s and 1 $b$ results in a $a^2 b$ product.
In fact, this type of analysis can be construed to be a demonstration of the binomial theorem, re for $n \in \mathbb{Z^+},$
$$(a + b)^n = \sum_{k=0}^n \binom{n}{k}a^k b^{(n-k)}.$$
The binomial theorem can alternatively be demonstrated through straight algebra, based on the idea that
$$\binom{n}{k} + \binom{n}{k+1} = \binom{n + 1}{k + 1}.$$
Addendum
Responding to the OP's request to re-check his query.
The image shown matches the binomial theorem.
Is it that no ways to arrange for the 3rd term since it would be C(2,1) . it is no ways to arrange them would be ab or ba.
If this is your question, then no. This is the wrong way to consider it. The easiest way to demonstrate how you calculate the coefficient of each term in the binomial expansion is to consider $(a + b)^4$, which can be represented as
$(a + b) ~\times$
$(a + b) ~\times$
$(a + b) ~\times$
$(a + b).$
Consider the terms:
$(a^4)$
There are $\binom{4}{4}$ ways of choosing 4 $a$'s from the 4 factors above.
$(a^3b)$
There are $\binom{4}{3}$ ways of choosing 3 $a$'s from the factors above.
$(a^2b^2)$
There are $\binom{4}{2}$ ways of choosing 2 $a$'s from the factors above.
$(ab^3)$
There are $\binom{4}{1}$ ways of choosing 1 $a$ from the factors above.
$(b^4)$
There are $\binom{4}{0}$ ways of choosing 0 $a$'s from the factors above.
Addendum-1
Responding to the OP's last comment.
If there 2 a’s , then there are also 2b’s.That means no of ways to arrange them is $\frac{4!}{2!}{2!}.$
Okay, now, I see what you are saying. Instead of focusing on the fact that in the Addendum multiplication, you were choosing 2 of the 4 $a$'s, you are focusing on the fact that if you have 2$a$'s and 2$b$'s and 4 slots to drop them in, there are $\binom{4}{2}$ ways of choosing which slots to assign to the $a$'s.
There is a bijection between arranging the 2$a$'s and 2$b$'s into 4 slots on the one hand, and choosing which of the 4 factors will contribute the 2$a$'s on the other hand.
Therefore, if I have understood you correctly, yes, you can explain the coefficient of any term $a^kb^{(n-k)}$ by asking how many ways can you select $k$ slots out of $n$ to position the k$a$'s.
On
In light of your addition edit here is a second answer:
It is misleading and unfortunate that the text used "arrangement". I believe a better (albeit semantically confusing and challenging) way of saying it would be "choices to position 2 out 3 items".
The coefficient of the $ab^2$ term in $(a+b)^3$ is ${3\choose 2} = 3$ is not because we are arranging $2$ out of three different objects, but instead it is because we are choosing out of $3$ copies of $b$ we are picking $2$ of them.
.....
Bear with me.
When we have $(a+b)^k$ we have $k$ copies of the pair $a,b$. (I'll illustrate with color)
$$(a+b)^k = \color{blue}{(a+b)}\color{red}{(a+b)}\color{green}{(a+b)}\color{orange}{(a+b)}$$
When we expand that out we will pick one $a$ or $b$ from each pair to get $2^k$ terms:
$$ \color{blue}{(a+b)}\color{red}{(a+b)}\color{green}{(a+b)}\color{orange}{(a+b)}=\\ \color{blue}a\color{red}a\color{green}a\color{orange}a+\color{blue}a\color{red}a\color{green}a\color{orange}b+\color{blue}a\color{red}a\color{green}b\color{orange}a+\color{blue}a\color{red}a\color{green}b\color{orange}b+\\ \color{blue}a\color{red}b\color{green}a\color{orange}a+\color{blue}a\color{red}b\color{green}a\color{orange}b+\color{blue}a\color{red}b\color{green}b\color{orange}a+\color{blue}a\color{red}b\color{green}b\color{orange}b+\\ \color{blue}b\color{red}a\color{green}a\color{orange}a+\color{blue}b\color{red}a\color{green}a\color{orange}b+\color{blue}b\color{red}a\color{green}b\color{orange}a+\color{blue}b\color{red}a\color{green}b\color{orange}b+\\ \color{blue}b\color{red}b\color{green}a\color{orange}a+\color{blue}b\color{red}b\color{green}a\color{orange}b+\color{blue}b\color{red}b\color{green}b\color{orange}a+\color{blue}b\color{red}b\color{green}b\color{orange}b$$
Now as a term with $m$ $b$s (and the remaining $k-m$ all being $a$s) are all the same and it doesn't matter which of the $k$ choices the $m$ $b$s came from we can combine then.
So the coefficient of the $a^{k-m}b^m$ will be the number of ways that, out of $k$ total copies of our $b$s, we can chose $m$ of them.
Let's draw a picture (For the picture I'll have to acknowledge I'm using $k = 4$).
$$\color{blue}a\color{red}a\color{green}a\color{orange}a+\color{blue}a\color{red}a\color{green}a\color{orange}b+\color{blue}a\color{red}a\color{green}b\color{orange}a+\color{blue}a\color{red}a\color{green}b\color{orange}b+\\ \color{blue}a\color{red}b\color{green}a\color{orange}a+\color{blue}a\color{red}b\color{green}a\color{orange}b+\color{blue}a\color{red}b\color{green}b\color{orange}a+\color{blue}a\color{red}b\color{green}b\color{orange}b+\\ \color{blue}b\color{red}a\color{green}a\color{orange}a+\color{blue}b\color{red}a\color{green}a\color{orange}b+\color{blue}b\color{red}a\color{green}b\color{orange}a+\color{blue}b\color{red}a\color{green}b\color{orange}b+\\ \color{blue}b\color{red}b\color{green}a\color{orange}a+\color{blue}b\color{red}b\color{green}a\color{orange}b+\color{blue}b\color{red}b\color{green}b\color{orange}a+\color{blue}b\color{red}b\color{green}b\color{orange}b=$$
[Ways to have $0$ $b$s:] $\color{blue}a\color{red}a\color{green}a\color{orange}a+$ [(${4\choose 0}=1$ ways to do it]
[Ways to have $1$ $b$:] $\color{blue}a\color{red}a\color{green}a\color{orange}b+\color{blue}a\color{red}a\color{green}b\color{orange}a+\color{blue}a\color{red}b\color{green}a\color{orange}a+\color{blue}b\color{red}a\color{green}a\color{orange}a+$ [(${4\choose 1}=4$ ways to do it]
[Ways to have $2$ $b$s:] $\color{blue}a\color{red}a\color{green}b\color{orange}b+\color{blue}a\color{red}b\color{green}a\color{orange}b+\color{blue}a\color{red}b\color{green}b\color{orange}a+\color{blue}b\color{red}a\color{green}a\color{orange}b+\color{blue}b\color{red}a\color{green}b\color{orange}a+\color{blue}b\color{red}b\color{green}a\color{orange}a+$ [(${4\choose 2}=6$ ways to do it]
[Ways to have $3$ $b$s:] $\color{blue}a\color{red}b\color{green}b\color{orange}b+\color{blue}b\color{red}a\color{green}b\color{orange}b+\color{blue}b\color{red}b\color{green}a\color{orange}b+\color{blue}b\color{red}b\color{green}b\color{orange}a+$ [(${4\choose 3}=4$ ways to do it]
[Ways to have $4$ $b$s:] $\color{blue}b\color{red}b\color{green}b\color{orange}b=$ [(${4\choose 4}=1$ ways to do it]
$${4\choose 0}a^4 + {4\choose 1}a^3b + {4\choose 2}a^2b^2 + {4\choose 1}a^3b + {4\choose 0}b^4=\\ a^4 + 4a^3b+ 6a^2b^2 + 4ab^3 +b^4$$.
And that's the binomial theorem!
......
so $$(a+b)^2 =\\\color{blue}{(a+b)}\color{red}{(a+b)}=\\\color{blue}a\color{red}a+\color{blue}a\color{red}b+\color{blue}b\color{red}a+\color{blue}b\color{red}b=\\\color{blue}a\color{red}a + \color{blue}b\color{red}b + (\color{blue}a\color{red}b+\color{blue}b\color{red}a)=\\a^2 + b^2 + 2ab$$
====
P.S. Neat little fact that comes out of all this meshugana is that ${n\choose 0} + {n\choose 1} + {n\choose 2} + ....... +{n\choose n-1} + {n\choose n} = 2^n$.
Who would have thunk it?!
Okay, I see what your book is getting at.
${3\choose 2}$ is the coefficient of $ab^2$. That is one $a$ and two $b$s. So you need the number of ways to arrange $a,b,b$. They are not three different things.
There are three ways to do it. they are 1) $a,b,b$ and 2) $b,a,b$ and 3) $b,b,a$.
There are two ways to think of ${3\choose 2}$. It can be thought of as the number of ways to pick two items out of three different items. (That's how I think of it and that is why the operation is call "$3$ CHOOSE $2$). But you book is thinking of it as: how many ways are there to arrange two different types of items if you have $2$ of one type, and the remaining out of $3$ items are all the second type.
Those two ways of thinking are equivalent because if you think of it the second way, all you are really doing is choosing $2$ out of three positions to place the first type of item.
=========
${3 \choose 2}$ is not the number of ways to arrange $A,B,C$.
${3\choose 2}$ is the number of ways to choose two elements from $A,B,C$. And the three ways are: 1) The two elements you choose are $A, B$. 2) the two things you choose are $A,C$ or 3) the two things you choose are $B,C$. Notice that choosing $A$ and $B$ are the exact same two things as choosing $B$ and $A$....
The number of ways to arrange $A,B,C$ are $3!=6$. The six ways to arrange are i)$ABC$, ii) $ACB$, iii) $BAC$, iv) $BCA$, v) $CAB$, vi) $CBA$. The number of ways to choose two items is a completely different question.
${2\choose 1}=2$ is the number of ways to choose ONE item from two. So the number of ways to choose one item from $A,B$ are two ways, 1) You can choose $A$, or 2) You can choose $B$. Those are the only two ways to choose one item from $A,B$.
=============
The binomial Thereom states if you expand out $(a + b)^k$ you will get a sum with $k+1$ terms and the coefficient the $j$th term will be ${k\choose j-1}$.
So the three terms when you expand $(a+b)^2$ will be $a^2 + 2ab + b^2$ (note the order of the coefficients). The first coefficient of the $a^2$ term will be ${2\choose 0} = 1$. There is one way to choose $0$ items from $2$ items that one way is to choose no items.
The second coefficient of the $ab$ term will be ${2\choose 1} = 2$. There are $2$ ways to choose one item from $2$ items. (You can chose one... or the other).
And the third coefficient of the $b^2$ term will be ${2\choose 2} = 1$. There is $1$ way to choose two items from $2$ items. That one and only way is to choose both the items.
So if we expand $(a+b)^2$ we get $(a+b)^2 = {2\choose 0}a^2 + {2\choose 1}ab + {2\choose 2}b^2 = 1\cdot a^2 + 2\cdot ab + 1\cdot b^2 = a^2 + 2ab +b^2$.
And if we expand $(a+b)^3$ we get $(a+b)^3 = {3\choose 0}a^3 + {3\choose 1}a^2b + {3\choose 2}ab^2 + {3\choose 3}b^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
Do you see why ${3\choose 0} = 1$? The only $1$ way to choose zero items is not to choose any.
Do you see why ${3\choose 1} =3$. The $3$ ways to choose one item, is to choose one item, and there are three to choose from.
Do you see why ${3\choose 2} =3$? There are three pairs we can choose.
Do you see why ${3\choose 3} =1$? The $1$ way to choose three items is to choose all of them.