Hey.
We started to study all this subject of combinatorics integrated with the subject of functions.
1. I don't actually understand how to integrate between combinatorics and function, those functions which represent our possibilities and etc...
And why at all we need to represent our combinatoric problem with an answer integrating functions?
What does that even mean? Integration (at least in the sense of calculus) involves continuous functions. Combinatorics is about discrete functions. This question appears to make no sense at all.
2. Secondly, we get this formula to calculate some sorts of possibilities:
\(\displaystyle \frac{n!}{(n-k)!} \)
It's not clear to me why is this really correct, what is posed behind it?
Correct for doing WHAT? You cannot seriously expect us to tell you why a formula is effective if you can't be bothered to say what it is supposed to be effecting.
3. For the Newton's binom - the lecturer has presented the proof for it but I don't understand it well, he also used the formula from above (2) to present the proof.
Binomial theorem or binomial coefficient?
Thank you!
Please take the time to formulate your questions carefully, and please pose one question per thread.
Because I have no clue what you are even remotely talking about in your first question, I shall mostly deal with question 2, which I suspect deals with permutations of n objects taken k at a time.
Let's make this concrete. You have five balls that can be distinguished by color: red (r), white (w), purple (p), green (g), and black (b). You have two boxes distinguished by size: large (L), and small (S). How many distinct ways can you have one of those 5 balls in each of the 2 boxes.
1 Lr, Sw
2 Lr, Sp
3 Lr, Sg
4 Lr, Sb
5 Lw, Sr
6 Lw, Sp
7 Lw, Sg
8 Lw, Sb
9 Lp, Sr
10 Lp, Sw
11 Lp, Sg
12 Lp, Sb
13 Lg, Sr
14 Lg, Sw
15 Lg, Sp
16 Lg, Sb
17 Lb, Sr
18 Lb, Sw
19 Lb, Sp
20 Lb, Sg
There are \(\displaystyle 20 = \dfrac{120}{6} = \dfrac{5!}{3!} = \dfrac{5!}{(5 - 2)!}.\)
Well the formula works in that case. Is that a fluke?
Let's think more generally. Suppose we have n distinguishable balls and k distinguishable boxes, where
\(\displaystyle k,\ n \in \mathbb Z \text { and }1 \le k \le n.\)
If n > 0 and k = 1, we can choose any one of n balls to put in box 1. Our number is n.
If n > 1 and k = 2, we can choose any one of n balls for the first box, but for each of those we can choose any of (n - 1) balls for box 2.
Our number is n(n - 1).
If n > 2 and k = 3, we can choose any one of n balls for the first box, but for each of those we can choose any of (n - 1) balls for box 2, and for each of those n(n-1) ways, we can choose (n - 2) balls for box 3.
Our number is n(n - 1)(n - 2).
In general then,
\(\displaystyle x = \text { the number of distinguishable ways to put one from a set}\)
\(\displaystyle \text {of n distinguishable balls into each of a set of k distinguishable boxes } = \displaystyle \left ( \prod_{j=n - k + 1}^n j \right ).\)
How does that formula work for our example of 5 balls and 2 boxes. It means
\(\displaystyle \displaystyle \left ( \prod_{j=n - k + 1}^n j \right ) = \left ( \prod_{j=5 - 3 + 1}^5 j \right ) = \left ( \prod_{j=4}^5 j \right ) = 4 * 5 = 20.\)
And we know that is the right answer.
Now if you do not understand the general case, try some more concrete ones. You might try six different colored balls and three boxes. (This will take you quite a bit of time. Don't leave any out.)
Good to here?
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) = \left ( \prod_{j=n-k+1}^n j \right ) * 1 \implies\)
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) =\left ( \prod_{j=n-k+1}^n j \right ) * \dfrac{(n - k)!}{(n - k)!} \implies\)
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) = \left ( \prod_{j=n-k+1}^n j \right ) * (n - k)! \div (n - k)! \implies\)
\(\displaystyle \displaystyle \left (\prod_{j=n-k+1}^n j \right ) = \left ( \prod_{j=n-k+1}^n j \right ) * \left ( \prod_{j=1}^{n-k} j \right ) \div (n - k)! \implies\)
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) = \left ( \prod_{1}^n j \right ) \div (n - k)! \implies\)
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) = n! \div (n - k)! \implies\)
\(\displaystyle \displaystyle \left ( \prod_{j=n-k+1}^n j \right ) = \dfrac{n!}{(n - k)!}.\)
