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MATHCOUNTS TOOLBOX Facts, Formulas and Tricks

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Lesson 10: Combinations

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When different orderings are not to be counted separately, i.e. the outcome, mn is equivalent to the outcome nm, the problem involves combinations.

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Combination Formula: Different orders of the same items are not counted. The combination formula is equivalent to dividing the corresponding number of permutations by r!. n: number of available items or choices r: the number of items to be selected Sometimes this formula is written: C(n,r).

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Taking the letters a, b, and c taken two at a time, there are six permutations: {ab, ac, ba, bc, ca, cb}. If the order of the arrangement is not important, how many of these outcomes are equivalent, i.e. how many combinations are there?

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Taking the letters a, b, and c taken two at a time, there are six permutations: {ab, ac, ba, bc, ca, cb}. If the order of the arrangement is not important, how many of these outcomes are equivalent, i.e. how many combinations are there? ab = ba; ac = ca; and bc = cb The three duplicate permutations would not be counted, therefore three combinations exist

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Calculate the value of 7 C 4.

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Calculate the value of 7 C 4. This represents a combination of 7 objects taken 4 at a time and is equal to

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Calculate the value of 9 C 5

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Calculate the value of 9 C 5 This represents a combination of 9 objects taken 5 at a time and is equal to...

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In how many ways can three class representatives be chosen from a group of twelve students? If the order of the arrangement is not important, how many outcomes will there be?

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In how many ways can three class representatives be chosen from a group of twelve students? If the order of the arrangement is not important, how many outcomes will there be? This represents a combination of 12 objects taken 3 at a time and is equal to

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Fini!

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