Flashcard 6 - Numbers and Operations Flashcard Set for the Math Basics


The correct answer is:

For permutations, the order matters. For combinations, order doesn’t matter.


Imagine the problem: “10 students are competing in a spelling bee. How many ways can students win gold, silver, and bronze?” You’d use a permutation for this one:

\[P(10, 3) \;\text{or}\; P\,^{10}_3 =\frac{10!}{(10-3)!} = \frac{10!}{7!}= 720\]

Now imagine changing the problem slightly: “Ten students are competing in a spelling bee. The top 3 students each win a blue ribbon. How many ways can this happen?” You’d use a combination for this one:

\[C(10,3) = \frac{10!}{(10-3)! \cdot 3!}=\frac{10!}{7! \cdot 3!}=120\]

To illustrate the difference, imagine Alice, Bob, and Charlie are the top 3 students. When the top three are ordered (gold, silver, bronze), there are many options: ABC, ACB, BAC, BCA, CAB, and CBA. This is a permutation. But when the order doesn’t matter (blue ribbons), then all of the 6 options are effectively the same.

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