Probability of a sequence of playing cards in a specific order

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I've just seen The Story of God with Morgan Freeman and in the last episode a psychology professor says that drawing six cards from a 52 card deck has only a probability of one in 14 billion to be in that exact order.

I understand that the probability of getting these six cards can be calculated like this: $$\frac{1}{\dbinom{52}{6}}$$ but how do I calculate the probability of them being in exactly one order.

My thought is, that for the first spot there is a 1/6 probability, for the second a 1/5 probability and so on. The problem is I don't get how I should put this thought into an equation with the other equation.
Thanks for your help in advance.

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There is one way to pick six particular cards in order.

How many ways can we select six of the $52$ cards in order?

We have $52$ options for the first card, $51$ options for the second card, $50$ options for the third card, $49$ options for the fourth card, $48$ options for the fifth card, and $47$ options for the sixth card. Hence, there are $$52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 = \frac{52!}{46!} = \binom{52}{6}6!$$ ordered selections of six cards, so the desired probability is $$\frac{1}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47} = \frac{1}{14~658~134~400}$$