We have three choices for the first person, two for the second and only one for the last. Handpicked Content: Contemplating Six Sigma and Internal Control Ignore that for a moment, let’s just figure out how many ways we can rearrange three people. Wait a minute! Andy/Bob/Charlie = Charlie/Bob/Andy. Either way, they’re all going to be equally disappointed. If I give a participant ribbon to Andy, Bob and Charlie, it’s the same as giving a participant ribbon to Charlie, Andy and Bob. How many ways can I award three participant ribbons to five people? Let’s say that instead of awarding blue, red and yellow ribbons for our barbecue contest, we award the top three with “participant” ribbons. The world looks the same to them whether it’s ordered or not ordered. You can mix the order up, and they’re still happy. They just go about their lives as if nothing really matters.Ĭombinations are the happy-go-lucky cousins of permutations. We all have a relative that is laid back. P(n,k) = n! / (n – k)! Combinations: No Order Needed In more general terms, if we have n items total and want to pick k in a certain order, we get:Īnd this is the permutation formula: The number of ways k items can be ordered from n items: This is saying, “use the first three numbers of 5!” If we divide 5! by 2!, we get: 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60 (because the 2 × 1 in the numerator and denominator will cancel each other out).Ī better (simpler) way to write this would be: 5! / (5 – 3)! Handpicked Content: Article Spotlight: Helping Your Customer Improve Processes What do we call 2 × 1? 2-factorial! This is what is left over after we pick three winners from five contestants. How do we get the factorial to “stop” at 3? We need to get rid of the 2 × 1. We only want 5 × 4 × 3 (the total number of options). To do this, we started with all five options then took them away one at a time (four, then three, etc.) until we ran out of ribbons.įive-factorial (written 5!) is: 5! = 5 × 4 × 3 × 2 × 1 = 120.īut 120 is too big! It would work if we had five ribbons. We had to order three people out of five. The total number of options was 5 × 4 × 3 = 60. All that matters is that we understand that we had five choices at first, then four and then three. Let’s say C wins the yellow ribbon.įor this example, we picked certain people to win, but that doesn’t really matter.
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