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Combination and Permutation 强化训练Tutorial Combination An arrangement of r objects, WITHOUT regard to ORDER and without repetition, selected from n distinct objects is called a combination of n objects taken r at a time. The number of such combinations is denoted by   The differ...

Combination and Permutation 强化训练
Tutorial Combination An arrangement of r objects, WITHOUT regard to ORDER and without repetition, selected from n distinct objects is called a combination of n objects taken r at a time. The number of such combinations is denoted by   The difference between combinations and permutations is in combinations you are counting groups (order is not important) and in permutations you are counting different ways to arrange items with regard to order. The n and the r mean the same thing in both the permutation and combinations, but the formula differs. Note that the combination has an extra r! in its denominator. Permutation An ORDER of arrangements of r objects, without repetition, selected from n distinct objects is called a permutation of n objects taken r at a time, and is denoted as   In other words, when you need to count the number of ways you can arrange items where ORDER is important, then you can use permutation to count. For example, you may want to know how many ways to pick a 1st, 2nd, and 3rd place winner from 10 contestants. Since you are arranging them in order, you could use a permutation to do this. Or if you were wanting to know how many ways your committee could pick a president, a vice president, a secretary, and a treasurer, you could use permutations.   Example 2: In how many ways can 8 CD’s be arranged on a shelf?   Since we are arranging these CD’s that means order is important. So we can use permutations to help us out here. First we need to find n and r : n is the number of CD's we have to choose from. What do you think n is in this problem? If you said n is 8 you are correct!!! There are 8 CD's in this problem. r is the number of CD’s we are using at a time. What do you think r is? If you said r is 8, pat yourself on the back!! We are arranging all 8 CD's on the shelf.   Putting this into the permutation formula we get:   *n = 8, r = 8 *0! = 1 *Expand out 8!     If you have a factorial key, you can put it in as 8! divided by 0! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 40320 different ways to arrange those 8 CD’s, that’s a lot.   Example 3: If a softball league has 10 teams, how many different end of the season rankings are possible? (Assume no ties).   Since we are ranking these teams that means order is important. So we can use permutations to help us out here. First we need to find n and r : n is the number of teams we have to choose from. What do you think n is in this problem? If you said n is 10 you are correct!!! There are 10 teams in this problem. r is the number of teams we are ranking at a time. What do you think r is? If you said r is 10, pat yourself on the back!! We are ranking all 10 teams.   Putting this into the permutation formula we get:   *n = 10, r = 10 *0! = 1 *Expand out 10!     If you have a factorial key, you can put it in as 10! divided by 0! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 3,628,800 different ways to rank those 10 teams, that’s a lot.   Example 4: In how many ways can a sorority of 20 members select a president, vice president and treasury, assuming that the same person cannot hold more than one office.   Since we are choosing offices, which is a way to rank members, that means order is important. So we can use permutations to help us out here. First we need to find n and r : n is the number of members we have to choose from. What do you think n is in this problem? If you said n is 20 you are correct!!! There are 20 members in this problem. r is the number of members we are selecting for offices at a time. What do you think r is? If you said r is 3, pat yourself on the back!! There are 3 offices.   Putting this into the permutation formula we get:   *n = 20, r = 3 *Expand 20! until it gets to 17! ( which is the ! in den) *Cancel out 17!'s     If you have a factorial key, you can put it in as 20! divided by 17! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 6840 different ways to select the three officers, that’s a lot.   Example 5: How many different arrangements can be made using two of the letters of the word TEXAS if no letter is to be used more than once?   Since we are arranging letters, this means order is important. So we can use permutations to help us out here. First we need to find n and r : n is the number of letters we have to choose from. What do you think n is in this problem? If you said n is 5 you are correct!!! There are 5 letters in TEXAS. r is the number of letters we are using at a time. What do you think r is? If you said r is 2, pat yourself on the back!! We are using 2 letters at a time.   Putting this into the permutation formula we get:   *n = 5, r = 2 *Expand 5! until it gets to 3! ( which is the ! in den) *Cancel out 3!'s     If you have a factorial key, you can put it in as 5! divided by 3! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. This means there are 20 different 2 letter arrangements.     Example 1: In a conference of 9 schools, how many intraconference football games are played during the season if the teams all play each other exactly once?   When the teams play each other, order does not matter, we are counting match ups. For each game there is a group of two teams playing. So we can use combinations to help us out here. Note that if we were putting these teams in any kind of order, then we would need to use permutations to solve the problem. But in this case, order does not matter, so we are going to use combinations. First we need to find n and r : If n is the number of teams we have to choose from, what do you think n is in this problem? If you said n = 9 you are correct!!! There are 9 teams in the conference. If r is the number of teams we are using at a time, what do you think r is? If you said r = 2, pat yourself on the back!! 2 teams play per game.   Let's put those values into the combination formula and see what we get:   *n = 9, r = 2 *Eval. inside ( ) *Expand 9! until it gets to 7! which is the larger ! in the den. *Cancel out 7!'s     If you have a factorial key, you can put it in as 9!, divided by 7!, divided by 2! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 36 different games in the conference.   Example 2: You are going to draw 4 cards from a standard deck of 52 cards. How many different 4 card hands are possible?   This would be a combination problem, because a hand would be a group of cards without regard to order. Note that if we were putting these cards in any kind of order, then we would need to use permutations to solve the problem. But in this case, order does not matter, so we are going to use combinations. First we need to find n and r : If n is the number of cards we have to choose from, what do you think n is in this problem? If you said n = 52 you are correct!!! There are 52 cards in a deck of cards. If r is the number of cards we are using at a time, what do you think r is? If you said r = 4, pat yourself on the back!! We want 4 card hands.   Let's put those values into the combination formula and see what we get:   *n = 52, r = 4 *Eval. inside ( ) *Expand 52! until it gets to 48! which is the larger ! in the den. *Cancel out 48!'s     If you have a factorial key, you can put it in as 52!, divided by 48!, divided by 4! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 270,725 different 4 card hands.   Example 3: 3 marbles are drawn at random from a bag containing 3 red and 5 white marbles. Answer the following questions (a - d):   3a. How many different draws are there?   This would be a combination problem, because a draw would be a group of marbles without regard to order. It is like grabbing a handful of marbles and looking at them. Note that there are no special conditions placed on the marbles that we draw, so this is a straight forward combination problem. Note that if we were putting these marbles in any kind of order, then we would need to use permutations to solve the problem. But in this case, order does not matter, so we are going to use combinations. First we need to find n and r: If n is the number of marbles we have to choose from, what do you think n is in this problem? If you said n = 8 you are correct!!! There are 3 red and 5 white marbles for a total of 8 marbles. If r is the number of marbles we are drawing at a time, what do you think r is? If you said r = 3, pat yourself on the back!! 3 marbles are drawn at a time.   Let's put those values into the combination formula and see what we get:   *n = 8, r = 3 *Eval. inside ( ) *Expand 8! until it gets to 5! which is the larger ! in the den. *Cancel out 5!'s     If you have a factorial key, you can put it in as 8!, divided by 5!, divided by 3! and then press enter or =. If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one. Wow, this means there are 56 different draws.   3b. How many different draws would contain only red marbles?   This would be a combination problem, because a draw would be a group of marbles without regard to order. It is like grabbing a handful of marbles and looking at them. In part a above, we looked at all possible draws. From that list we only want the ones that contain only red. Let's see what the draw looks like: we would have to have 3 red marbles to meet this condition: 3 RED First we need to find n and r : If n is the number of RED marbles we have to choose from, what do you think n is in this problem? If you said n = 3 you are correct!!! There are a total of 3 red marbles. If r is the number of RED marbles we are drawing at a time, what do you think r is? If you said r = 3, pat yourself on the back!! 3 RED marbles are drawn at a time.   Let's put those values into the combination formula and see what we get:  
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