Abracadabra!

Theorem: It is impossible to have exactly nine cards arranged in order. Proof: For the sake of contradiction, assume that exactly nine cards are arranged in order. Then the last card must be in order as well, as it has no other place to go (or else the original nine cards are not in order). This would mean that all ten cards are in order, a contradiction to the original assumption. Thus $p=0$ , and our desired answer is $\boxed{0}$ .

By "in order" I assume that a hand such as $10,1,2,3,4,5,6,7,8,9$ would work - no wrap arounds First we pick $9$ of the integers in $\dbinom{10}{9}$ ways. There is only one permutation of them which satisfy the conditions Then we have to pick where the last integer can go - normally, it would have $10$ spots but one of them will violate the conditions; all $10$ cards would be in order. So there are $9$ spots that the last integer can go. Finally, there are $10!$ ways to order the cards without regard to the restrictions. The probability $p = \dfrac{\dbinom{10}{9} \cdot 9}{10!} = \dfrac{1}{8!}$ Then $1000p = \dfrac{1000}{8!} < 1 \implies \lfloor 1000p\rfloor = \boxed{0}$

P is 0 because you cannot have just one card in the incorrect spot. The least number of cards that can be placed in the incorrect spot is two.

Now, since P=0,

So, 1000P=1000*0=

                          0

It's impossible to obtain an arrangement in which exactly nine of the ten cards are in order. Either all cards are in order, or else at least two of the cards must not be in order, leaving at most $10-2=8$ cards in order.

Assume 9 of the cards are in order. There is only one remaining place for the last card to be, but that place would put the card in order. Therefore it is impossible to have exactly 9 cards in order. Thus the probability is 0,

If nine of the cards are already in order, in their numerical positions, the last one has only one place to go. That position corresponds to the numerical value of the card and makes all 10 cards in order. Thus there is no way that exactly 9 are in order. $p = 0$

If you start with 10 cards and exactly 9 cards are in order, then all the 10 cards are in order. So the probability of exactly 9 cards being in order is 0.

There are $10!$ possible ways for the cards to be ordered. However only in two of them exactly 9 of the cards are arranged in numerical order. They are:

10, 1,2,3,4,5,6,7,8,9

and

2,3,4,5,6,78,9,10 ,1

Thus the probability that exactly 9 of the cards are arranged in numerical order would be $\frac{2}{10!}$ . Consequently:

$\lfloor{1000\times\frac{2}{10!}} \rfloor = \frac{5}{9 \times 8 \times 7 \times 6 \times 3} = 0$

It is not possible for only 1 card to be in a wrong position. For a card to be in a wrong position, it has to occupy the supposed position of another card, and that card will in turn be displaced to another wrong position. Therefore, p=0, and 1000p=0.

If exactly 9 of the cards are arranged in order, then no matter what, the 10th card must be filled in the last spot, so all 10 cards must be arranged in order. This is a contradiction since we assumed that exactly 9 cards were arranged in order. So the probability that exactly 9 are in order is 0, and 0x1000 = 0, our answer.

If 9 cards are in order,then the 10th card will also be in order.Thus the probability that exactly 9 cards are in order is 0 .

The total ways of arranging the cards=10!=3628800

The number of ways that exactly 9 of the cards are arranged in order=1x10=10

p = the probability that exactly 9 of the cards are arranged in order = The number of ways that exactly 9 of the cards are arranged in order/The total ways of arranging the cards=10/3628800=2.7557E-6

1000p=1000x2.7557E-6=2.7557E-3

The floor value of 1000p is thus 0