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What is the probability that, pulling out 7 cards from a deck of $40$ different cards, without reinserting the cards in the deck, there are $3$ cards that I wrote on a piece of paper in advance (that is, there are $3$ cards chosen in advance)?

My book says: $\frac{7}{1976}$

I've tried to solve the probability problem as follows:

There are disp $(40, 7) = 40*39*38*37*36*35*34=93963542400$ different ways to create subsets of $7$ cards from a deck of $40$ different cards.

Leaving out the three preselected cards, there are disp $(37, 4)=37*36*35*34=1585080$ different ways to select $4$ cards from a deck of $37$ different cards.

The three preselected cards can be estracted in 3! = 6 different ways, so I have:

$6 * \frac{1585080}{93963542400} = \frac{0,2}{1976}$ but the reference solution gives $\frac{7}{1976}$

Who is right? Who is wrong?

Thank you in advance for considering my request.

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    $\begingroup$ You are on the right track, but the one thing you forgot to do is choose the 3 places out of 7 where your preselected cards will go, which can be done in $\binom{7}{3}=35$ ways. If you multiply your answer by 35, I think you will get the correct answer. (Using combinations, as in the answer below, is easier, though.) $\endgroup$ – user84413 Sep 10 '14 at 22:25
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The numbers you are using are for selecting cards in a particular order. I have not fully analyzed your numbers to find the error in your reasoning, but here is a solution.

There are $\binom{40}{7}=\frac{40!}{7!(40-7)!}$ ways of selecting seven cards from a deck of 40, assuming that we do not care about the order the cards are drawn in. How many ways are there to draw seven cards such that all 3 of our marked cards are taken? We have to take the 3 marked card, and then choose 4 from what remains (note that, if we cared about order, we would NOT be able to choose the marked cards separate from the non-marked cards like this, as we would have to look at all the ways the three marked cards could be inserted into the other cards). There are $\binom{40-3}{7-3}=\binom{37}{4}=\frac{37!}{4!(37-4)!}$ ways to choose the 4 remaining cards. The probability we are looking for is this

$$ \binom{37}{4}/\binom{40}{7} = \frac{37!}{40!}\frac{7!}{4!}\frac{33!}{33!} =\frac{7\cdot 6 \cdot 5}{40\cdot 39\cdot 38}=\frac{7}{1976}$$

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    $\begingroup$ Aaron, thanks so much for having helped me! Answer accepted, thanks again! ;-) $\endgroup$ – Massimiliano Bottegoni Sep 11 '14 at 3:39
  • $\begingroup$ Fun fact: as long as you have $k$ card to draw with $m$ draws from $n$ cards, you get the probability not only by $$\frac{\binom{n-k}{m-k}}{\binom{n}{m}}$$ but also by $$\frac{\binom{m}{k}}{\binom{n}{k}}$$. This may seem like simple arithmetic, but historically the second was the one I discovered first back in school :) $\endgroup$ – SK19 May 7 '18 at 16:04

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