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formula for permutation and combination
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From Encyclopedia
permutations and combinations
permutations and combinations see probability .
From Yahoo Answers
Question:So my math teacher(geometry) asked us to define a LONG list of math vocabulary. Many were quite simple, and I got right away. Some of these, I've never heard of. I can't find good explanations of them online, and I don't have a textbook handy. Anything you know would be appreciated!
(note: am supposed to give examples too so I need to really understand how they work)
The vocab words are as follows:
combinations
distance formula
function notation
inverse operations(algebraic)
midpoint formula
permutations
subsets
Thanks! :)
Answers:Here are some hints on where to find things. Wikipedia is a good starting place. In most cases they will have an article about the term. Another good trick is to search Google for define:[term you are interested in] without the [ ] Like define:permutation subsets can be found on pages about set theory a set is a "bunch" of objects. A subset is none, some, or all of them, but not anything else from outside the set. permutations and combinations go together. so a search on both terms at once should turn up relevant pages (Permutations are ordered subsets, combinations are unordered subsets.) distance formula is related to the Pythagorean theorem. midpoint formula is related: it's the point half the distance between the ends. inverse operations are opposites inverse of addition is subtraction inverse of multiplication is division function notation  not sure about this one, I would think of f(x) "eff of x", but not sure how it's defined. =
Answers:Here are some hints on where to find things. Wikipedia is a good starting place. In most cases they will have an article about the term. Another good trick is to search Google for define:[term you are interested in] without the [ ] Like define:permutation subsets can be found on pages about set theory a set is a "bunch" of objects. A subset is none, some, or all of them, but not anything else from outside the set. permutations and combinations go together. so a search on both terms at once should turn up relevant pages (Permutations are ordered subsets, combinations are unordered subsets.) distance formula is related to the Pythagorean theorem. midpoint formula is related: it's the point half the distance between the ends. inverse operations are opposites inverse of addition is subtraction inverse of multiplication is division function notation  not sure about this one, I would think of f(x) "eff of x", but not sure how it's defined. =
Question:Six people are to sit at a circular table. Two of the people are not to sit immediately
beside each other. Find the number of ways that the six people can be seated.
This is my shot:
The total number ways that people can be seated is
6 p 6 = 720
The ways that those two of the people will sit near each other is
6 p 2 = 30
Therefore, the answer is 690.
Any ideas?
Answers:since a round table doesn't have a defined starting point, formula for # of permutations is (n1)! as against n! for a row for this particular problem, imagine that there are only 4 chairs seat the 4 people who don't have any restrictions the first person can be seated anywhere, [this becomes the "starting point" henceforth ] the remaining 3 can be seated in 3P3 = 3! = 6 ways there are 4 spaces betwee the 4 seated where chairs can be inserted we need to insert 2, so 4P2 = 4*3 = 12 ways multiply the two to get 6*12 = 72 ways 
Answers:since a round table doesn't have a defined starting point, formula for # of permutations is (n1)! as against n! for a row for this particular problem, imagine that there are only 4 chairs seat the 4 people who don't have any restrictions the first person can be seated anywhere, [this becomes the "starting point" henceforth ] the remaining 3 can be seated in 3P3 = 3! = 6 ways there are 4 spaces betwee the 4 seated where chairs can be inserted we need to insert 2, so 4P2 = 4*3 = 12 ways multiply the two to get 6*12 = 72 ways 
Question:1.) 128 tennis players are entered in a tournament. If the top 32 players are seeded, how many ways could we assign the seeds to 32 different players?

2.) 100 Senators are being considered to chair 17 committees in the US senate. How many different ways could the chair people be selected, assuming that no one chairs more than one committee?

3.) 8 friends go to the movies and Mike insists on being third in line to get tickets. If his wife always stands in front of him in line, how many ways can the 8 friends line up at the ticket window?

4.) 10 students wish to go to lunch. How many ways could we line up 5 of them, send them to lunch, then line up the remaining 5?

5.) A bookshelf contains 3 math, 3 science and 3 social studies textbooks. If all of the books are different, how many ways can the books be arranged on the shelf?

6.) How many ways can the books be arranged such that the books of the same type are together?

7.) A baseball team has 25 players on its roster. How many ways are there to pick 9 starters?

8.) The local bank branch has a pool of 8 tellers and 8 customer service reps. How many ways can the manager select 4 tellers and 2 customer service reps to work on a given day?

9.) Pingpong balls labled 115 are placed in a bag. How many ways can I reach into the bag and pull out either four or six balls?

There are 10 people at a party, and each pair of people shakes hands exactly once. How many handshakes were there?

How many subsets of a 15element set have either four or six elements?
Answers:hi jenny ! no one is going to solve so many q's for you. if u really want help, u shd post only ones u can'tsolve, at most 2 in one post. anyway, i'll solve possibly the most difficult q of the lot for u q3 mike & his wife are standing at fixed places in the line that leaves 6 others, who can be arranged in 6P6 ways= 6! =720 ways  just remember, order doesn't matter in a combination, but it does in a permutation you should be able to do the q's on ur own. they aren't too tough. good luck, sorry i saw ur q so late
Answers:hi jenny ! no one is going to solve so many q's for you. if u really want help, u shd post only ones u can'tsolve, at most 2 in one post. anyway, i'll solve possibly the most difficult q of the lot for u q3 mike & his wife are standing at fixed places in the line that leaves 6 others, who can be arranged in 6P6 ways= 6! =720 ways  just remember, order doesn't matter in a combination, but it does in a permutation you should be able to do the q's on ur own. they aren't too tough. good luck, sorry i saw ur q so late
Question:I need to know wether the problem is a permutation or combination. Then I also need the formula. Thank you! And if you can please explain how you got if its a permutation or combination. Thanks again!
1. Five students from the 90 students in your class not running for class president will be selected to count the ballots for the vote for class president. In how many ways can the 5 students be selected?
2. 20 students are running for 3 different positions on student council. In how many ways can the 3 positions be filled?
3. To complete a quiz, you must answer 3 questions from a list of 6 questions. In how many ways can you complete the quiz?
4. The buyer for a retail store must decide which sweaters to stock for the upcoming fall season. A sweater from one manufacturer comes in 5 different colors and 3 differeent textures. The buyer decides that the store will stock the sweater in 3 different colors and two different textures. How many different sweaters are possible?
Thanks again!
Answers:1. Since it does not matter in which order they are chosen, C(90,5) = 90 89 88 87 86/5 4 3 2 1 = 5273912160/120 = 43,949,268 2. Since the positions are presumably different, P(20,3) = 20 19 18 = 6840 3. Snce the order in which you answer does not matter, C(6,3) = 6 5 4/3 2 1=20 4. C(5,3) C(3,2) = 10 3 = 30
Answers:1. Since it does not matter in which order they are chosen, C(90,5) = 90 89 88 87 86/5 4 3 2 1 = 5273912160/120 = 43,949,268 2. Since the positions are presumably different, P(20,3) = 20 19 18 = 6840 3. Snce the order in which you answer does not matter, C(6,3) = 6 5 4/3 2 1=20 4. C(5,3) C(3,2) = 10 3 = 30
From Youtube
Letters Game Solver  Permutations and Combinations :Download: www.thinkfn.com Permutations: We can count the number of possible joint choices by using the factorial function: n' = n*(n1)*(n2)*...2*1 Excel formula = PERMUTAR(n;x) The factorial function grows faster than exponential (like n^n.) In some crossword/letters games we need to choose a subset of a larger set (ex: Choosing a word with 5 letters from six and arranging them on a shelf. If you fortget the order for a moment, the only you need is the number of ways to choose. This is called combinations. You can use the Pascal's triangle (see wikipedia) or you can use Microsof Excel to compute these values: Excel formula = combin(n,x)
Permutations and Combinations :A video looking at the basic definitions of permutations and combinations.