Probability of an event happening on the first attempt What is the probability that when selecting numbers out of a hat, with 10 people and numbers 1-10, how likely is it that the 3 people who aren't present for this event, are selected in the 8, 9 and 10 spot?
 A: It helps to rather than selecting people in the usual order of the first spot first, the second spot second, and so on... to instead pick the people in the opposite order.  The original number we pick out of the hat will be the person put in the tenth spot, the next number we pick out will be the person put in the ninth spot, and so on.
Assuming we don't care which of the three absent people was specifically the tenth, which was the ninth and so on, just so long as the three people are among the three final spots in any order, we get a probability of: 
$$\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}=\frac{6}{720}=0.00833\overline{3}$$
This could have been efficiently described also using binomial coefficients and the hypergeometric distribution as $\dfrac{\binom{7}{0}\binom{3}{3}}{\binom{10}{3}}$, again yielding the same result as before.
A: I'm going to go out on a limb here and ask for the following clarification: 
Are you really asking how likely is it that the 7 present people colluded against the absent 3, perhaps because you are suspicious of your fantasy football league's draft order?  
If so, the answer is more complicated than simply the one in 120 probability that JMoravitz gave.  Instead, we need to use Bayesian inference.  
Suppose, prior to knowing the outcome of the drawing, you suspect collusion will happen with probability $p$, and that if collusion occurs, the colluders will certainly put the absentees at the end.  By Bayes' rule, the probability that the outcome occurred via collusion, as opposed to naturally is $$\frac{120p}{119p+1}$$  Assigning fair priors is hard, especially after you witness an event such as this.  For prior probabilities for $p$ even as low as 1%, this leads to a better than coin flip chance that collusion occurred.
EDIT, POST OP COMMENT: How likely it is that your friends colluded depends on what you think $p$ is, ie how likely you thought those 7 friends were to collude BEFORE the drawing took place.  Are your friends very honest people and do you think the chances of them colluding are 1 in a million?  If so, then $p=.000001$, and your revised guess is about 1 in 9000, so they probably didn't collude, even though the results seem fishy.  If you thought there was a 25% chance of collusion before the draft, then upon seeing the results you should think there's a 97.5% chance of collusion.  Your semi-shady friends seem very guilty in the light of these results.  
The formula above is merely guide that says, if you would have expected collusion with probability $p$ before, then now that you have seen the results, you should think the chances that they colluded are $\frac{120p}{119p+1}$.  I don't know you or your friends, so I can't really guess what $p$ is reasonably.  
A: There are $$10!=10\times9\times8\times 7\times\cdots\times2\times 1$$
possibilities (possible orders)to draw the ten numbers without replacement. But if three places are already occupied then there are only
$$7!= 7\times\cdots\times2\times 1$$
possibilities.
So the probability at stake is
$$\frac{7!}{10!}=\frac{1}{10\times9\times8}=\frac1{720}.$$
EDIT
It seems that the OP wanted one fixed order of the people numbered $8, 9, 10$. If this order is not fixed (you can mix the numvers withinn the chosen positions) you have to multly my result by $$3!=6.$$
