Source. |

One technique-able counting problem type that you might come across on the SAT is what I’ll call a “possibilities” problem*. It might involve cards (but *not* playing cards – the SAT doesn’t like those), or pictures being lined up on a wall in different orders. Your job will be to determine the number of possible outcomes given a particular scenario. Like so:

- Mike is arranging seven of his various awards and commendations on a shelf in his office. If he insists that his hard-fought Class of 1999 Math Award be placed in the center, in how many different orders could he arrange the seven items?

(A) 60

(B) 72

(C) 120

(D) 720

(E) 1440

The best way to tackle a possibilities problem like this is to draw a bunch of blanks like you’re about to play Hangman, and then start thinking, methodically, through the choices you have at every step along the way. I’m going to illustrate this process with slightly more thoroughness than you probably will on test day (you won’t need to make up award names, but I will because it’s *hilarious*):

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Class of 1999 Math Award | ||||||

Choices | 1 |

First, as I did above, you must account for any special conditions or restrictions. The Math Award must go in the middle, so there’s *only one choice* for Position 4.

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Invisible Man Award | Class of 1999 Math Award | |||||

Choices | 6 | 1 |

Once all the restrictions are accounted for, start filling in the rest of the spaces. To fill Position 1, Mike has 6 different awards to choose from. Say, for argument’s sake, he chooses the Invisible Man Award next.

Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Award | Invisible Man Award | Acne League – Most Improved | Push Ups Contest – Last Place | Class of 1999 Math Award | Spin the Bottle – Luckiest Player Ever | 5^{th} Grade Science Fair – 2^{nd} Place | Shortest Fight in School History – Loser |

Choices | 6 | 5 | 4 | 1 | 3 | 2 | 1 |

To fill the next spot, since he used up the Invisible Man Award, he has 5 awards left to choose from. Once he chooses the Acne League – Most Improved award for Position 2, he has 4 choices for Position 3. He continues this process until he’s filled all the positions.

To calculate the number of different arrangements Mike could have made, multiply the number of choices he had at every step:

6 × 5 × 4 × 1 × 3 × 2 × 1 = 720

**When solving a possibilities problem, set up the hangman blanks, then imagine yourself actually performing the task described. First take care of special conditions or restrictions, and then take care of everything else. At every step, ask yourself “How many choices do I have?****” And then ask yourself “How many choices do I have now?” And then – you guessed it – ask yourself how many choices you have ****again****. **

**Stop when you run out of choices.**

##### Possibly you’re interested in more practice?

You need to be registered and logged in to take this quiz. Log in or Register

*Aside from this footnote, I’m deliberately avoiding the terms “permutation,” “combination,” and “factorial” here. That’s not because I don’t know them; it’s because I’ve found that they manufacture more confusion than they alleviate on the SAT. Remember: The SAT is not a math test! If you prefer to solve the questions laid out here by cramming them into nPr and nCr notations in your calculator, be my guest, but don’t cry to me when you miss counting questions on the SAT.