At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following problem:

King Louis XIII must pick a team of 5 musketeers to investigate one of Cardinal Richelieu’s nefarious schemes. If there are 10 musketeers to choose from, what is the probability that four of them (Athos, Porthos, Aramis, and d’Artangan, of course) will be selected?

A) 1/2

B) 1/10

C) 3/5

D) 1/42

E) 5/252

To solve this problem, we must first remember that the probability of any event is calculated by taking the number of desired outcomes over the number of possible outcomes. In this case, figuring out the number of desired outcomes is not too difficult. We know who four of the five musketeers should be, so the only variable is the remaining musketeer. We have already used 4 out of the 10 possible musketeers, so there are 6 possibilities left for the remaining musketeer. If we let P, Q, R, S, T, and U represent the unknown musketeers, then we could represent the desired outcomes like so:

Athos, Porthos, Aramis, d’Artangan, and P

Athos, Porthos, Aramis, d’Artangan, and Q

Athos, Porthos, Aramis, d’Artangan, and R

Athos, Porthos, Aramis, d’Artangan, and S

Athos, Porthos, Aramis, d’Artangan, and T

Athos, Porthos, Aramis, d’Artangan, and U

That leaves figuring out the total number of possible outcomes. You could try to write down all the possible combinations of five musketeers, but with 10 musketeers to choose from that’s going to take a long time, and there would be many opportunities for making mistakes. What we are trying to figure out here is how many possible *combinations* of 5 musketeers we could make from a group of 10. To calculate this, all we need is a little formula that you might remember from math class:

Where *n* is the number of items to choose from and *r* is the number of items to be selected. Combinations and permutations are occasionally tested on the SAT, so you would do well to memorize this formula and other relevant formulas before test day. Using the formula, we find that the total number of ways to select a group of 5 from a group of 10 is:

## 252

Thus, the number of desired outcomes over the number of possible outcomes is:

Thus 1/42, choice D, is correct. If you know what to do, it takes only about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until then, keep up the good work and happy studying!

Your problem is still confusing. If you have 1 person to pick out of 6, how did you get the 42?

Hi Juanita, since this probability, we first need to know the total number of possibilities or combinations possible. Because we have 10 musketeers and are making a group of 5, we’ll have 10C5, or 252 possible combinations. We then know we must choose the four listed musketeers, meaning it doesn’t matter who the last of the five is. That means we have six distinct possibilities out of 252 that satisfies what the questions asks for (i.e. must have Athos, Porthos, Aramis, and d’Artangan, along with an inconsequential fifth member), giving us 6/242 or 1/42. Let us know if you still have questions about this!