this is the 27 on destroyer
'there are 12 members of the Democratic party at the meeting, how many groups of 3 member can be formed?
it turned out to be this is combination.
89 on destroyer
'Find the number of ways six different object can be tken three at time'
and this is permutation..
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can someone explain why one is permutation and one is not?
and I just can't get the idea of permutation and combination..
is there any good books or website I can learn this concept?
I will teach this to you in one quick post and you can save time not reading a book. I hope
Combination means order doesn't matter while permutation means order does matter.
In the first you just want to know how many ways you can select 3 people out of 12. It doesn't matter which one you pick first so you use a combination and get nCr = 12!/3!9! =12*11*10 / 6 = 220.
In the second you want 6 objects taken 3 at a time. But this is a poorly worded question so don't try to learn anything from it. I would see this being a combination as well because it seems like you just need to select 3. There's no reason why one object being selected before another should matter. I'd say the answer is nCr = 6!/3!3! = 20 but they apparently want nPr = 6!/3! = 120.
Combination: Order doesn't matter.
Example: You are running a contest where there are 10 finalists. You randomly choose 3 winners from a jar. Each prize is $1,000. How many ways can this be done?
Since all 3 prize levels are the same it doesn't matter which person is picked first, second, or third. So there are nCr = 10!/3!7! = 120 ways to pick the 3 people.
Permutation: Order matters.
Example: You are running a contest where there are 10 finalists. You randomly choose 3 winners from a jar. First prize is $5,000. Second prize is $2,500. Third prize is $1,000. The first name drawn gets third prize, the second name drawn gets second prize, and the third name drawn gets first prize. How many ways can this be done?
Since the 3 prize levels are different, it DOES matter who gets picked first. So there are nPr = 10!/7! = 720 ways to pick the 3 people.
Theory: Factorial means multiply all natural numbers equal to and less than the starting number. So n! = n*(n-1)*(n-2)*...*2*1. Hopefully you know this.
Start by asking how many ways can you seat 5 people in 5 chairs in a row? You can choose any of the 5 for chair #1, any of the remaining 4 for chair #2, etc. This is just 5! = 120. That's how many ways you can pick out 5 people from a group in a certain order. But what if you only want 3 of them?
Well you can choose any of the 5 for spot #1, any of the remaining 4 for spot #2, and any of the remaining 3 for spot #3. So that's 5*4*3. But is there a way to write this mathematically for any number of people?
Normally you have 5! = 5*4*3*2*1. You want just 5*4*3. But how do you get rid of the 2*1? Well, you can divide by 2!. This leaves you with 5*4*3. Similarly, if you have 10! and want to select only 4 people, you want to have 10*9*8*7. You want the 6*5*4*3*2*1 to not be there. So you take 10! and divide by 6!.
In the most general sense, if you let r = the number of people you want to select out of a group of n people, you can do this in n!/(n-r)! ways. The (n-r)! term leaves you with n(n-1)(n-2)...(n-r+1) which is what you want.
Now how about combinations which pay no attention to order? Notice in the above that any of the 10 people could be chosen first, then any of the remaining 9 could be chosen next, etc. This is a permutation because it matters which of the 10 people we chose first, which of the 9 people we chose next, etc. How many ways, though, can the 4 people which we chose be chosen?
In other words, we selected 4 people out of 10. Let's say the 10 people were ABCDEFGHIJ. Let's say I chose ABCD. I could choose them that way. I could choose them in the order CADB. I could choose them in the order BADC. In a combination we don't care how we choose them. We just care that person A, person B, person C, and person D were all picked. Who cares how?
What if I chose BCDE? I could choose them in the order ECBD, etc. Notice that for any UNIQUE group of 4 people, there are a certain number of ways I could have selected them. How many ways can I select 4 people? Well... that'd be 4!. I could have picked any of the 4 of them first, any of the remaining 3 second, etc. So for any given group of 4 people, there will be 4! = 24 ways to choose them. So knowing that 10!/(10-4)! = 5,040 = the number of 4 person groups where order matters, there are 5,040/24 = 210 UNIQUE 4 group selections that don't pay attention to order (WHY?).
So what I did there was take the formula for nPr and divided by r!. The formula for nPr is n!/(n-r)! so the formula for nCr is [n!/(n-r)!] / r! which equals n!/(n-r)!r!
