How many possible combinations of 12 numbers are there, such that the first six numbers are ascending and the last

combinatoricsdiscrete mathematicsprobability

Given the numbers 1,2,3,…, 12. How many possible combinations of ordering these numbers are there, such that the first 6 numbers and the last six numberes are ascending?

May someone please give me a hint on how to compute the solution? I know, there are 12! possible combinations of these 12 numbers and $\binom{12}{6}$ possibilities of choosing 6 numbers out of 12. But how do I eliminate those of not ascending order?

Best Answer

There are $\binom{12}{6}$ ways of choosing which six numbers will appear first. There is only one way to arrange them to be ascending. The remaining six numbers will appear last, and there is only one way to arrange them to be ascending.