In how many ways can you order the numbers: 1-10 in line so that to the right of 6 there are only numbers larger than 6.

Question

First, I wanted to order 1-5 : $10 \choose 5$

Then I wanted to place 6 to the right of the 1-5 order and then to order the 4 larger number.

My answer was $10 \choose 5$$\cdot$ 4! , but the textbook answer is $10!\over6$

Answer 1

Let $k$ be the amount of numbers greater than $6$ that are to its left. Then the solution is:

$$ \sum_{k=0}^4 {4 \choose{k}} (5+k)! (4-k)! = 5!4! + 4\cdot6!3! + 6 \cdot 7!2! + 4\cdot 8! + 9! = 604'800 = \frac{10!}{6}. $$

Answer 2

Consider the collection of permutations, there are $10!$ of them and consider where the six number $\{1,\cdots 6\}$ might appear in the permutation (ignore the other numbers). In any given permutation, each of $\{1,\cdots, 6\}$ is equally likely to appear last. Thus exactly $\frac 16$ of the permutations have $6$ last amongst those $6$.