N is a natural number N>1, and K is a natural number between 1 and N. In how many permutations of numbers 1 to N ... N is a natural number N>1, and K is a natural number between 1 and N. In how many permutations of numbers 1 to N the number K is smaller then all the numbers  on his right?
The answer says: C(n,k)(k-1)!(n-k)!=n!/k
But why C(n,k) appears? C(n,k) says that we choose places for 1,..,k numbers from 1,..,n possible place, however in this case the restriction allows 1,..,k to be only at the beginning because all the number on k's right must be bigger...so why still C(n,k) appears? For example if 1,2,3,4,5 then  C(n,k) will allow 4,5,1,2,3
 A: First pick the $n-k$  places where $k$ and all the numbers $\textbf{smaller}$ than $k$ are going to be in $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ ways. Clearly $k$ must be in the rightmost of the $k$ places. So permute the places to the left in any of $(k-1)!$ ways. Finally: permute the numbers larger than $k$ in $(n-k)!$ ways. So the answer is $\binom{n}{k}(n-k)!(k-1)!=\frac{n!}{(k)!(n-k)!}(n-k)!(k-1)!=\frac{n!}{k}$
A: 1) First choose the places in which the integers $\{k+1, k+2, \cdots, n\}$ will go;
this can be done in $\;\;\;\binom {n}{n-k}=\binom{n}{k}$ ways.
2) Next put these integers in order in their places, which can be done in $(n-k)!$ ways.
3) Next place the integer k in the first place from the right which is not filled.
4) Finally, place the integers $\{1,\cdots,k-1\}$ in order in the remaining places, which can be done in $\;\;\;(k-1)!$ ways.
The number of ways to do this is given by $\displaystyle\binom{n}{k}(n-k)!(k-1)!=\frac{n!}{k}$.
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A similar, but slightly simpler argument, is given by the following:
1) First choose the places in which the integers $\{1, \cdots, k\}$ will go;
this can be done in $\binom{n}{k}$ ways.
2) Next put k in the last of the places chosen.
3) Then put the elements of $\{1,\cdots,k-1\}$ in order in the other places chosen, which can be done in $\;\;\;(k-1)!$ ways.
4) Finally, place the elements of $\{k+1,\cdots,n\}$ in order in the remaining places, which can be done in $\;\;\;(n-k)!$ ways.
The number of ways to do this is given by $\displaystyle\binom{n}{k}(k-1)!(n-k)!=\frac{n!}{k}$.
A: Let's say that $N=8$ and $K=5.$
The $5$ must be in one of the last four spots.
So with $5$ in the fifth spot, the numbers to the right are $6,7,8$ in any order ($3!$ permutations).  The numbers to the left can also be in any order ($4!$ permutations).  Also note here that there is only one way to choose the numbers that are to the right:  $3 \choose 3$.  Total for this case is $4!3! {3 \choose 3}.$
Now, with $5$ in the sixth spot, we can have $6,7$ or $6,8$ or $7,8$ in the last two spots, and they can be in any order.  The five numbers to the left can also be in any order, so for this case we have $5!2!{3 \choose 2}$ cases.
For the $5$ in the seventh spot, we can have $6,7,$ or $8$ in the last spot.  Similarly we have $6!1!{3 \choose 1}$ cases.
For $5$ in the last spot, we have $7!0!{3 \choose 0}$ cases.
So the total becomes
$$\sum_{x=0}^{N-K}x!(N-x)!{{N-K} \choose x}=8064, (N=8, K=5).$$
This agrees with the answer you have: ${8 \choose 5}(5-1)!(8-5)! = 8!/5 = 8064.$
(If I can get the two formulas to agree, I'll edit my answer, but this is as far as I've gotten.)
A: The numbers $1,2,\dots,k$ are treated symmetrically when writing down all permutations. So, the probability of $k$ being first among them is $\dfrac 1k$.
Therefore, the number of permutations with that property is $\dfrac{n!}k$.
