For many GRE test takers combinations and permutations represent one of the great challenges of the quantitative section. After all, who can remember what the technical distinctions are between those two terms and what is up with those excited numbers followed by an exclamation point?? Well, good news everybody, you too can excel at determining how many possible ways something can be done without those complex nCr or factorial (x!) notations by following the simple steps we'll outline in this post! Although, you probably should know that 5! simply means 5 x 4 x 3 x 2 x 1 for future possible reference in a possible GRE problem

When Order Matters
Let's consider a simple GRE style question - How many ways can the gold, silver, and bronze medals be awarded to the top three of 10 competitors in a swim meet?

Step 1
To begin, write out a line as a spot for each selection to be made labeling the known restrictions. In this case use shorthand to indicate Gold, Silver and Bronze:
G: __
S: __
B: __

Step 2
Next, insert the number of available options at the moment of selection into each of the open spots for selection considering if repetition can occur.
So, for the Gold slot there would be 10 available options for selection so fill in as G: 10
Then, for the Silver slot repetition would not be allowed, so fill in S: 9
Lastly, for the Bronze slot repetition would once again not be allowed, so fill that in as B: 8

Step 3
Multiply those values in the open spots straight across since no cancellation of equivalent outcomes is required to find that the number of ways that the Gold, Silver and Bronze medals can be awarded in order to 10 competitors in a swimming competition would be 10 x 9 x 8 = 720 possible outcomes.

When Order Doesn't Matter
But, what about when order doesn't matter? Well, we need to make sure we don't double count equivalent outcomes and we can do that with one simple change to the prior setup! So, let's consider a slight variation to our prior question - How many ways can each of the top three of 10 competitors in a swim meet be awarded a medal of honor?

Step 1
Once again, write out a line as a spot for each selection to be made labeling the known restrictions. In this case use shorthand to label the choices using MH for Medal of Honor:
MH: __
MH: __
MH: __

Step 2
Then, insert the number of available options at the moment of selection into each of the open spots for selection considering if repetition can occur.
So, for the first Medal of Honor slot there would be 10 available options for selection so fill in as MH: 10
Next, for the second Medal of Honor slot repetition would not be allowed, so fill in MH: 9
Lastly, for the third Medal of Honor slot repetition would once again not be allowed, so fill that in as MH: 8

Step 3
Now, because of the possibility of equivalent outcomes, order is said to not matter and if order doesn't matter you must divide the number of options we've already placed by the number of equivalent outcomes at the moment of selection to cancel out double counting.

When the first selection is made there are three available equivalent outcome MH spots for the first competitor, so our first medal of honor spot must be divided by three. Place this as MH: 10/3
Then, after the first selection is made only two equivalent MH spots remain for the second selected competitor. Place this as MH: 9/2
Finally, at the moment of the third selection only one spot remains open, so place this for consistency sake as MH: 8/1

Step 4
With the fractions of \(\frac{10}{3}x\frac{9}{2}x\frac{8}{1}\) cancel the 9 in the second numerator with the 3 in the first denominator as well as the 10 in the first numerator with the 2 in the second denominator before processing the now simplified multiplication operation of 5 x 3 x 8 = 120 ways that three identical medals of honor can be awarded to 10 competitors in a swimming competition.

As you can see the approach is really similar for both when order does and when order doesn't matter with the big difference of whether to divide or not. So, simplify the consideration by asking two questions for yourself:
1) Does Order Matter?
2) Do I Divide?
If the answer to the first question is Yes, the answer to the second question is no.
If the answer to the first question is No, the answer to the second question is yes.

Apply this approach and you'll never need to solve complex combinatorics problems on the GRE with an excited number again!