Hi,
I am having a lot of trouble trying to distinguish between when I should use the 5!/4!3! form versus 5!/4! for instance...in other words, combinations v permutations. I read the chapter a couple of times but still feel like I don't have a good grasp on the differences. I know that permutations relate to order but a lot of the questions in the problem set on pg 61 that use permutation give no indication that order is relevant.
I.e. problems #13 and 3 use permutation. What is the indicator, however, that order matters?
Another issue I"m having is with problem#9. In the answer, the word is 12345N. I don't quite understand what is meant by the use of the numbers. I'm assuming the numbers represent factorials we exclude in the denominator. But why?
Thanks for your help.
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- sarahme
- rfernandez
- Course Students
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- Joined: Fri Apr 07, 2006 8:25 am
You're correct that permutations are when the order of the selection matters and that combinations are when the order of the selection does not matter.
Traditionally, the way to go about these counting problems is to start there: Is this a permutation or a combination? If it's a permutation of n things taken r at a time, then you use the formula n!/(n-r)! If it's a combination, then you use the formula n!/[r!(n-r)!].
Here's an example of each:
3 people are to be chosen from a group of 5 people to serve on a committee. How many ways can groups of 3 be chosen?
Let's say the people are named A, B, C, D, and E. We can see that the committee comprised of members B, D, and E is no different from the committee comprised of E, D, and B. From this example, we can determine that the order of selection does not matter to this particular problem, so we use the combination formula, using n = 5 and r = 3.
n!/[r!(n-r)!]
5!/[3!(5-3)!]
5!/(3!2!)
10
Now another example:
5 people are swimming in a race. How many different 1st place, 2nd place, 3rd place outcomes are possible?
Again, let's call the people A, B, C, D, and E. Here, we see that if A, C, D finish first, second, and third respectively, then that's a different outcome from C, A, D finishing first, second, and third respectively. In other words, the order of the selection does matter to this problem. We use the permutation formula with n = 5 and r = 3.
n!/(n-r)!
5!/(5-3)!
60
The Strategy Guide employs a different method that is rooted in the same concepts as above, but explicitly uses anagrams. The method requires that you know how to find the number of anagrams for any given "word." Sometimes words have all distinct letters/symbols as in "BEACH" or "SIMPLE" or even "XYZ12". In some cases, words have repeated letters/symbols as in "LEVER" or "ASSESSES" or "11222NN".
To determine the number of anagrams possible for a word, you start by counting how many letters there are total. That number's factorial determines the numerator. Then, you count each instance of repeated letters. The product of these numbers' factorials determines the denominator. It's easier to see how this works through examples:
"SIMPLE" = 6! anagrams
Notice that there are six letters in the word, so 6! is the numerator. Since there are no repeated letters, there's nothing else to do.
"LEVER" = 5!/2! anagrams
There are five letters in the word, so 5! is the numerator. E is repeated twice, so 2! is the denominator.
"ASSESSES" = 8!/(5!2!) anagrams
There are eight letter in the word, so 8! is the numerator. S is repeated five times and E is repeated twice, so (5!2!) is the denominator.
Moving on, the anagram method takes a combinatorics problem and "translates" it into a word. What's interesting about the method is that the combination/permutation issue is essentially "baked" into the word that you construct. It's all about which symbols you use to describe the situation of the problem at hand. Let's answer the same two questions above using the anagram method.
3 people are to be chosen from a group of 5 people to serve on a committee. How many ways can groups of 3 be chosen?
Again, let's call the people A, B, C, D, and E. Now, for each possible committee, three will be on the committee (designated by Y for "yes") and two will not be on the committee (designated by N for "no"). One possible committee has members A, B, and C on it while D and E are not on it. It can be written up as "YYYNN." Another possible committee has A, D, and E on it while B and C are not on it. This committee can be described as "YNNYY." We can go on doing this, but essentially what we're trying to do is figure out how many anagrams the word "YYYNN" has. Well, fortunately, we have a method to determine the number of anagrams, as described above.
"YYYNN" = 5!/(3!2!) = 10
Again, there are five letters in the word, so 5! is the numerator. There are three Ys and 2 Ns in the word, so (3!2!) is the denominator.
5 people are swimming in a race. How many different 1st place, 2nd place, 3rd place outcomes are possible?
For this problem, we need new symbols to designate 1st, 2nd, and 3rd places. So we can use 1, 2, and 3. Otherwise, swimmers who come in after 3rd can be designated N for not coming in 1st, 2nd, or 3rd. So what's the anagram for this problem? "123NN"
"123NN" = 5!/2! = 60
As you can see, the results are identical but the means by which to get there are different. The benefit of the anagram method is that you essentially apply one clean, efficient method to both types of problems. But if the formulas work out better for you, then you should use that instead. I hope this is helpful.
Traditionally, the way to go about these counting problems is to start there: Is this a permutation or a combination? If it's a permutation of n things taken r at a time, then you use the formula n!/(n-r)! If it's a combination, then you use the formula n!/[r!(n-r)!].
Here's an example of each:
3 people are to be chosen from a group of 5 people to serve on a committee. How many ways can groups of 3 be chosen?
Let's say the people are named A, B, C, D, and E. We can see that the committee comprised of members B, D, and E is no different from the committee comprised of E, D, and B. From this example, we can determine that the order of selection does not matter to this particular problem, so we use the combination formula, using n = 5 and r = 3.
n!/[r!(n-r)!]
5!/[3!(5-3)!]
5!/(3!2!)
10
Now another example:
5 people are swimming in a race. How many different 1st place, 2nd place, 3rd place outcomes are possible?
Again, let's call the people A, B, C, D, and E. Here, we see that if A, C, D finish first, second, and third respectively, then that's a different outcome from C, A, D finishing first, second, and third respectively. In other words, the order of the selection does matter to this problem. We use the permutation formula with n = 5 and r = 3.
n!/(n-r)!
5!/(5-3)!
60
The Strategy Guide employs a different method that is rooted in the same concepts as above, but explicitly uses anagrams. The method requires that you know how to find the number of anagrams for any given "word." Sometimes words have all distinct letters/symbols as in "BEACH" or "SIMPLE" or even "XYZ12". In some cases, words have repeated letters/symbols as in "LEVER" or "ASSESSES" or "11222NN".
To determine the number of anagrams possible for a word, you start by counting how many letters there are total. That number's factorial determines the numerator. Then, you count each instance of repeated letters. The product of these numbers' factorials determines the denominator. It's easier to see how this works through examples:
"SIMPLE" = 6! anagrams
Notice that there are six letters in the word, so 6! is the numerator. Since there are no repeated letters, there's nothing else to do.
"LEVER" = 5!/2! anagrams
There are five letters in the word, so 5! is the numerator. E is repeated twice, so 2! is the denominator.
"ASSESSES" = 8!/(5!2!) anagrams
There are eight letter in the word, so 8! is the numerator. S is repeated five times and E is repeated twice, so (5!2!) is the denominator.
Moving on, the anagram method takes a combinatorics problem and "translates" it into a word. What's interesting about the method is that the combination/permutation issue is essentially "baked" into the word that you construct. It's all about which symbols you use to describe the situation of the problem at hand. Let's answer the same two questions above using the anagram method.
3 people are to be chosen from a group of 5 people to serve on a committee. How many ways can groups of 3 be chosen?
Again, let's call the people A, B, C, D, and E. Now, for each possible committee, three will be on the committee (designated by Y for "yes") and two will not be on the committee (designated by N for "no"). One possible committee has members A, B, and C on it while D and E are not on it. It can be written up as "YYYNN." Another possible committee has A, D, and E on it while B and C are not on it. This committee can be described as "YNNYY." We can go on doing this, but essentially what we're trying to do is figure out how many anagrams the word "YYYNN" has. Well, fortunately, we have a method to determine the number of anagrams, as described above.
"YYYNN" = 5!/(3!2!) = 10
Again, there are five letters in the word, so 5! is the numerator. There are three Ys and 2 Ns in the word, so (3!2!) is the denominator.
5 people are swimming in a race. How many different 1st place, 2nd place, 3rd place outcomes are possible?
For this problem, we need new symbols to designate 1st, 2nd, and 3rd places. So we can use 1, 2, and 3. Otherwise, swimmers who come in after 3rd can be designated N for not coming in 1st, 2nd, or 3rd. So what's the anagram for this problem? "123NN"
"123NN" = 5!/2! = 60
As you can see, the results are identical but the means by which to get there are different. The benefit of the anagram method is that you essentially apply one clean, efficient method to both types of problems. But if the formulas work out better for you, then you should use that instead. I hope this is helpful.
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