# Calculating permutations if the sequences have to be in ascending order?

How would you go about calculating the number of permutations in ascending order.

Obviously if you had (a set of) 3 numbers you have $3!$ permutations:

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1)

But only one of these is in ascending order (1,2,3).

Consider the lottery - picking 6 numbers from 49. Normally the order for this doesn't matter you just calculate '49 choose 6' but if the numbers have to be in ascending order how would you calculate that?

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In the same way: The number of (strictly) ascending sequences of $6$ numbers chosen from $49$ is $\binom{49}{6}$. For as you have pointed out, once the $6$ numbers have been chosen, there is precisely one way to arrange them in ascending order.