8 Choices, 4 spots = number of options?

There is a commercial airing right now about a fast food chain having 8 menu items to choose from, pick four. You can pick the same item 1, 2, 3, even 4 times. The commercial claims that there are 320 possibilities. How so? My numbers are huge...

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  • 1 decade ago
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    Let C(n,k) be the binomial coefficient n!/k!(n-k)! (C(n,k) is the number of ways to select k things from n things, when order doesn't matter). Clearly, you want to use C(n,k), since your meal doesn't depend on the order in which you choose the menu items. (In case you're unfamiliar with the notation, n! = n(n-1)(n-2)...1).

    The number of ways to pick 4 items is C(8,4) = 70

    The number of ways to pick 3 items is 3*C(8,3) = 168 (there are 3 ways in each choice to choose which item is doubled)

    The number of ways to pick 2 items where one item is tripled is 2*C(8,2) = 56.

    The number of ways to pick 2 items where each item is doubled is C(8,2) = 28.

    The number of ways to pick 1 item, quadrupled, is C(8,1) = 8.

    Adding these all together, the total is 330, so they were close. If the numbers you're getting are much bigger, it could be because you're counting the order in which the items are chosen.

    Hope this helps (& please don't hold me responsible for any errors in arithmetic -- it's late).

    Source(s): just me.
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