The Fundamental Counting Principle states that if an event has x possible outcomes and another independent event has y possible outcomes, then there are xy possible ways the two events could occur together. The toughest SAT probability questions involve counting, permutations and combinations. If you can do these difficult questions, you’re well on your way to better scores in SAT Math!
1. The probability of Jane getting an ice cream cone is 1/3. The probability of Jane buying a soda is ¼. What is the probability Jane will get an ice cream cone and buy a soda?
According to the Fundamental Counting Principle, we must multiply the two probabilities. The answer is 1/3 x ¼ = 1/12. Let’s look at a hard question:
2. How many three-digit integers have either 1 or 2 as their tens digit and 4 as their units digit?
To solve this problem, we need to find the possible outcomes for each digit (hundreds, tens, and units) and multiply them. Each digit was 10 possible numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9). The hundreds digit can be any of them except 0 (since a three-digit number cannot begin with 0). The tens digit was only 2 possibilities as stated in the question. The units digit has only 1 possibility. According to the Fundamental Counting Principle, the total number of possible numbers is 9 x 2 x 1 = 18. Notice how the numbers themselves don’t matter, only the number of possibilities.
Permutations are sequences. In a sequence, order is important.
3. How many different ways can five people stand in line?
For the first spot in line, we have five people to choose from, and as we go down the line trying to fill each spot, the number of people we have to choose from will decrease by 1. Therefore we have 5 x 4 x 3 x 2 x 1 = 120 ways
Some harder permutations problems may require you to use this formula:
n = the total number of options
r = the number of options chosen
The exclamation point is called a factorial. It’s a shorter way of writing the product of all positive integers less than or equal to the integer in front of the exclamation point.
For example, 4! = 4 x 3 x 2 x 1 = 24.
4. At Loma Linda Middle School’s track competition, 14 athletes are competing in the pole vault finals. How many possible options are there for the first 4 finishers?
Here n = 14 and r = 4. Since the order in which the athletes finish matters, we know to use the Permutation formula:
n! / (n – r)! = 14! / (14 – 4)! = 14! / 10! = 14 x 13 x 12 x 11 = 24, 024 options
Combinations are groups. Order doesn’t matter. The Combination formula is only slightly different from the Permutation formula:
n = total number of options
r = the number of options chosen
Notice the only difference is that the combination formula has an extra r! in the denominator.
5. Jose took 14 photos with her new digital camera. He wants to choose 10 of them to put on her Flickr profile. How many different groups of photos are possible?
Since the question asks about groups and not the order of the photos, we know this is a Combination problem.
n! / r! (n – r)! = 14! / 10! (14 – 10)! = 14! / 10! 4!
= 14 x 13 x 12 x 11 / 4 x 3 x 2 x 1 = 1,001 different groups
Always remember to ask yourself whether order matters to the problem you are trying to solve, and don’t forget the Fundamental Counting Principle! For more help with these challenging questions, schedule a 1-hour lesson with a Grockit tutor. Buy the lessons in packs of $50/hr on the Tutoring tab in the Grockit lobby, or purchase them individually at the time of the scheduled lesson.