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How can you prove the combination formula? - Answers
Combination Formula ProofGENERIC:Let C(n,r) be the number of ways to generate unordered combinationsThe number of ordered combinations (i.e. r-permutations) is P(n,r)The number of ways to order a single one of those r-permutations P(r,r)The total number of unordered combinations is the total number of ordered combinations (i.e. r-permutations) divided by the number of ways to order each combinationThus, C(n,r) = P(n,r)/P(r,r) = [n!/(n-r)!]/r!/(r-r)!] = n!/r!(n(n-r)!SPECIFIC:Let C(52,5) be the number of ways to generate unordered Poker handsThe number of ordered poker hands is P(52,5) = 311,875,200The number of ways to order a single poker hand is P(5,5) = 5! = 120The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each handThus, C(52,5) = P(52,5)/P(5,5)
Bing
How can you prove the combination formula? - Answers
Combination Formula ProofGENERIC:Let C(n,r) be the number of ways to generate unordered combinationsThe number of ordered combinations (i.e. r-permutations) is P(n,r)The number of ways to order a single one of those r-permutations P(r,r)The total number of unordered combinations is the total number of ordered combinations (i.e. r-permutations) divided by the number of ways to order each combinationThus, C(n,r) = P(n,r)/P(r,r) = [n!/(n-r)!]/r!/(r-r)!] = n!/r!(n(n-r)!SPECIFIC:Let C(52,5) be the number of ways to generate unordered Poker handsThe number of ordered poker hands is P(52,5) = 311,875,200The number of ways to order a single poker hand is P(5,5) = 5! = 120The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each handThus, C(52,5) = P(52,5)/P(5,5)
DuckDuckGo
How can you prove the combination formula? - Answers
Combination Formula ProofGENERIC:Let C(n,r) be the number of ways to generate unordered combinationsThe number of ordered combinations (i.e. r-permutations) is P(n,r)The number of ways to order a single one of those r-permutations P(r,r)The total number of unordered combinations is the total number of ordered combinations (i.e. r-permutations) divided by the number of ways to order each combinationThus, C(n,r) = P(n,r)/P(r,r) = [n!/(n-r)!]/r!/(r-r)!] = n!/r!(n(n-r)!SPECIFIC:Let C(52,5) be the number of ways to generate unordered Poker handsThe number of ordered poker hands is P(52,5) = 311,875,200The number of ways to order a single poker hand is P(5,5) = 5! = 120The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each handThus, C(52,5) = P(52,5)/P(5,5)
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- og:descriptionCombination Formula ProofGENERIC:Let C(n,r) be the number of ways to generate unordered combinationsThe number of ordered combinations (i.e. r-permutations) is P(n,r)The number of ways to order a single one of those r-permutations P(r,r)The total number of unordered combinations is the total number of ordered combinations (i.e. r-permutations) divided by the number of ways to order each combinationThus, C(n,r) = P(n,r)/P(r,r) = [n!/(n-r)!]/r!/(r-r)!] = n!/r!(n(n-r)!SPECIFIC:Let C(52,5) be the number of ways to generate unordered Poker handsThe number of ordered poker hands is P(52,5) = 311,875,200The number of ways to order a single poker hand is P(5,5) = 5! = 120The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each handThus, C(52,5) = P(52,5)/P(5,5)
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