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I need some help understanding how to approach problems with permutations/combinations.

Could someone first explain when I should be using combinations and when I should be using permutations?

Then could you help me work through some examples?

Example 1 : A standard deck of poker cards has 52 cards, consisting of 13 cards each in 4 suits. How many ways is it possible to draw five cards of the same suit? Consider the draw to be ordered.

I think that the answer is $$ 4 * \frac{13!}{8!} $$

Is this correct or am I completely off?

Example 2: For dinner at a favorite restaurant, there are four possible appetizers, five possible entrees, three possible desserts, and six possible beverages. How many ways are there to order a meal if you can select at most one of each, and you must choose an entree and a beverage, but the appetizer and dessert are optional?

I have no idea how to approach this example, any hints?

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  • $\begingroup$ For appetizer, we have $5$ choices, including no thanks. For every one of these choices of entree. For every one of the $5\times 5$ choices of appetizer and entree, we have $4$ choices about dessert. Continue. $\endgroup$
    – André Nicolas
    Oct 1, 2014 at 23:13

1 Answer 1

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You should use combinations when order doesn't matter and permutations when order does matter.

Example 1: First, you pick a suit, for which there are 4 choices. Once you have your suit, you're choosing five denominations of the 13 available. Hence there are $$4 \times \binom{13}{5}$$ flushes possible.

Example 2: We can include a "no appetizer" and "no dessert" as an option. Then, there are $$ (4+1) \times 5 \times (3+1) \times 6$$ possible course options.

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