3 selected cards on a set of 7 cards from a deck of 40 different cards. 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.
 A: 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
$$ \binom{37}{4}/\binom{40}{7} = \frac{37!}{40!}\frac{7!}{4!}\frac{33!}{33!} =\frac{7\cdot 6 \cdot 5}{40\cdot 39\cdot 38}=\frac{7}{1976}$$
