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This does not include putting the card back into the deck after it has been drawn.

I am comfortable with the number of ways two draw any two cards in a row (basic permutation of $52!/(52-2)!=52\times 51$.

I am also comfortable with the number of ways to draw any two spades in a row: $13!/(13-2)!=13\times12$ since there are 13 spades to choose from.

However, I am finding it hard to visualise and compute the way in which any card other than a spade is drawn, followed by a spade.

I think the correct answer is $39\times13=637$ from basic observation. (I.e. if there only existed 1 spade and 39 other cards then the number of permutations is 39, so with 13 spades the number of permutations is multiplied by 13.) But I want to do this in terms of well-defined permutation/combinatoric maths.

Thanks.

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    $\begingroup$ $39\times 13$ is indeed correct (note: there are 39 non-spades, not 49. I expect you had a typo). I highly recommend getting out of the habit of trying to look for "cure-all" formulas and feeling the need to write this in terms of factorials or binomial coefficients. Its simply not necessary. This follows from a much more fundamental principle: the rule of product. $\endgroup$ – JMoravitz Aug 31 '17 at 15:17
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    $\begingroup$ Huh. I never knew that name for it. I've always seen it called the "fundamental principle of counting", or something highfaluitn like that. $\endgroup$ – G Tony Jacobs Aug 31 '17 at 15:23
  • $\begingroup$ @JMoravitz Oh OK so I had it already. Don't worry I don't have that habit, which is why my first attempt at the problem using simple logic was successful... However, I strongly believe that if I come up with a solution then it is always good to generalise it for both future use and clarity. $\endgroup$ – ODP Aug 31 '17 at 15:32
  • $\begingroup$ @ JMoravitz not sure why you decided to place it as a comment where this is an excellent answer. $\endgroup$ – adhg Aug 31 '17 at 15:45
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    $\begingroup$ Note however that drawing two cards, one a non-spade, one a spade, can be done in $39\cdot 13 + 13\cdot 39$ ways. Specifying the ordering restricts to your answer. $\endgroup$ – Joffan Aug 31 '17 at 15:59
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You mean like this? $$V_{39,1}\cdot V_{13,1}=\dfrac{39!}{(39-1)!}\cdot \dfrac{13!}{(13-1)!}=39\cdot 13$$

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There are 39 non-spades in a deck and 13 spades. So you have 39 choices for your first card and 13 for the next one. That makes 39*13 choices.

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  • $\begingroup$ You just answered my question with the solution I provided in the question... (My 49 was quite clearly a typo.) $\endgroup$ – ODP Aug 31 '17 at 15:33
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    $\begingroup$ You're right, I shouldn't have skimmed your question. For what it's worth, counting problems are often best solved using this "basic observation" method. So I think what you did was the best way. $\endgroup$ – FullofDill Aug 31 '17 at 16:04

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