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It is quite confusing.....

2007-02-23 13:14:03 · 4 answers · asked by tralala 1 in Science & Mathematics Mathematics

4 answers

Just remember that there are MORE permutations than combinations. Just remembering the numbers for ONE example would suffice.

Or remember a phrase like "combinations of 5 things taken 3 at a time", which would make little sense if the meanings were reversed.

Or remember that this is also called "5 choose 3", and focus on how the letter "c" shows up in both cases.

2007-02-23 13:20:02 · answer #1 · answered by Curt Monash 7 · 0 0

combinations: sets, groups, committees, teams
permutations: arrangements, 3-digit numbers, 3-letter words

3-digit numbers is just an example, it can be any number of digits, the same goes for words.

ask yourself this question: in this particular example, does order matter? if it does, use permutations. if not, use combinations.

example: Al, Bob, Chris, and Dave run a race.
1) How many ways are there to pick 1st, 2nd 3rd place. Order matters since 1st is not the same as 2nd place. So use 4P3.
2) How many ways are there to pick a group of three people from the 4 runners. Since it does not matter which way we pick in a group, we use 4C3.

2007-02-23 13:50:28 · answer #2 · answered by Anonymous · 0 0

The key word is "order", which may or may not appear in the statement of the problem, but is generally implied.

Does the order of the items matter? If so, use permutations.

If the order doesn't matter, use combinations.

Does it matter whether the two people are chosen as president and vice president, in which case order matters, or are the two people simply a committee where neither position matters.

Generally the problem will suggest whether order matters.

2007-02-23 13:37:19 · answer #3 · answered by fcas80 7 · 0 0

the main isn't any count if order concerns. as an occasion, in card issues like "what share procedures are you able to pick 5 enjoying cards such which you have precisely 3 aces", then you definately do no longer care concerning to the order. the respond to that question is (4C3)(48C2). you utilize the mixtures like "4 pick 3" considering the fact which you prefer 3 out of the 4 aces and you do no longer care concerning to the order. for each decision of three aces, there are (48C2) ordinary procedures to pick the different 2 enjoying cards.

2016-12-14 04:18:27 · answer #4 · answered by libbie 4 · 0 0

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