How many ways are there to pick “n” objects from “n” objects if order does NOT matter?
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M$6 Answers
If the order does not matter, you have only 1 way to pick n objects out of n. This is because the only thing that matters is which ones you choose, not the order of that choice. Once you're done choosing, in any order, all n objects are chosen, hence there is only 1 way. The more general formula for picking k objects out of n (n greater than or equal to k) is: n!/(k! * (n-k)!) where 0! = 1. In this case, where k=n (your case), you have n!/(n! * 0!) = n!/n! = 1.
In all the above, the exclamation mark denotes the factorial function, where k-factorial, or k! = 1*2*...*k (if k=0, then k! = 1; if k=1, then k!=1; if k=2, then k! = 1*2=2).
If the order does matter, the answer is n! as was explained by @drmatt. This is because you have n choices for the first item (i.e. n possibilities to pick item #1), and for each of those choices you then have one fewer to choose from for the second slot, so you have to multiply the initial n by (n-1). For the third slot you have yet one fewer, or (n-2), so for the first three you have n*(n-1)*(n-2) possible picks. This continues until you are left with one slot and one item, so the overall number of possibilities is n*(n-1)*(n-2)*...*2*1 which if you reverse the order (as is allowed in multiplying whole numbers) gives you 1*2*...*n which is n!.
20 years of math and physics education, and 18 years of practice thereof.
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M$Now, @opher is the only correct one here thus far in your question. C(n,n) = 1. Simply put, if n = 2 and those things are A and B, then the orders of A then B and B then A count as the same thing. If you were asking for a permutation in which order is important, then the answer for n = 2, P(n, n) = 2.
However, there is one level further on this problem. Some people have mentioned reuse of the items. Lets say that each item can only be used once. Then you want to combine a of n items where a < n.
So the formula goes something like this C(n, a) = (n!)/(a! * (n-a)!) So lets look at n = 5 and a = 3, that is combining any 3 things from a group of five (and order does not matter).
(5!) / (3! * (5 - 3)!)
= 120 / (6 * 2!)
= 120 / (6 * 2)
= 120 / 12
= 10 combination!
However, if order matters, then the formula is different and it is called a permutation. Again, this formula uses a < n and no objects can be reused. (n!) / ((n - a)!)
(5!) / ((5 - 3)!)
= 120 / (2!)
= 120 / 2
= 60 permutations!
Finally, the reason there are 10 conminations and 60 permutations.
If you have 3 objects, then there are six possible order or these objects (3!). Thus each combination consists of six permutations. So 10 conminations * 6 possible orders = 60 permutations.
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M$
suppose you are making license plates with 3 letters and you can use the letters over and over, - order will not matter,
with 26 letters it would be 26 x 26 x 26
now suppose you could only use each number ONCE and had 3 spots, again order would not matter it would be 26 x 25 x 24.
The number of objects you have for the maximum decreases by one each time and you multiple according to how many picks you have.
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M$
26 letters in the alphabet --
pick up anyone off the table --
your first choice you have 26 available
next you can only pick 25, then 24.
n in this case is 26
n factorial is 26 * 25 * 24
n ! or n factorial is the correct answer -- and I'll go up and vote for the fist guy who gave that answer.
my source is my memory of high school or maybe early college math. probability and statistics books will back this up.
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M$n x (n-1) x (n-2) x (n-3) x ....... x {n-(n-1)} ways
Yeah! Order matters but the total number of ways remain the same. Order matters differently.
That's too short, but hope that helps because that is the actual answer.
Thanks
Academics
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M$
Yeah! You are right. But what about the outcome? Refer the example you used. When picked randomly, it is factorial n! (there, it is 6, right?). And when picked in order, the result is again factorial n! (again 6, right?).
When speaking of an order that involves all the objects, the answer will remain the same.
But if you specify an order where a particular object cannot be touched (counted) twice, then the answer is of course different.
Hope I am clear.
Thanks
I think the chief point being made here is that order does not matter as specified by the asker.
Looking at @omicron list of apple, orange and banana, it will make sense.
Lets say that you pick up the order apple, orange, banana.
Then dump the basket.
Finally, pick up the order of banana, orange, apple.
The combination of fruits picked up is the same both times. The order, aka permutation, is different. However, the asker specified that order does not matter. Therefore, anytime the basket contains an orange, apple, and banana in any order, it is counted as the same combination.
Thus when you have n object and you pick up all n of them, you can have only one combination, but n! permutations.
I think we almost understand each other, but just to be clear, the answer to the question as it is asked is 1, not n!.
There you are @ferg2kk! You got my point. That's what I said that the results would be same irrespective of the order of picking up the objects, because you are ultimately picking everything up. Isn't it?
Had it been specified otherwise (one object cannot be picked up twice), the results would have been different.
Thanks
What do you say about the outcome in the example of 3 objects used by @omicron? Aren't the results same?
Please give an example using an ordered pick Vs. random to prove mathematically that the results will differ. That will help me understand the thing.
Thanks
No. If the order doesn't matter, as @omicron said, you can tip the basket over and you get the only pick you can - all N objects. Thus the answer is 1, not N! (unless the order matters, or N=1).
If you have n objects, there are n! (n factorial) ways to pick from them.
For example... I have three fruit. How many ways can I pick the one that goes in the first "slot"? 3. I have two left for the second slot and that leaves 1 for the last slot. How many ways? 3x2x1 (3!) = 6.
Order can't matter... then you can't pick. If you have three fruit and you HAVE to have an apple first, then, there is no choice. The apple HAS to be first. (Unless you mean something different from "order" or you mean pick n objects from k objects).
You just take it one "slot" at a time. How many choices do you have first the first choice, then the second, then the third... etc. and multiply them together.
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M$
Just for completeness, in the case where order matters, the general formula for how many permutations you have for picking out k out of n objects (as explained by @ferg2kk) is:
n!/(n-k)! (which is equal to n*(n-1)*(n-2)*...*(n-k+1))
For example, if you have 26 objects, you're picking out any 3, and the order matters, then n=26, k=3, and the number of permutations (as in the example provided by @petpeople) would be:
26!/(26-3)! = 26!/23! = 26*25*24 = 15,600
This is the number of 3-letter "words" you can make up in the English language that have no repeating letters (whether they currently have meaning such as "and", or not, such as "nda").
In the specific case where n=k, the general formula becomes n!/(n-n)! = n!/0! = n! (because 0! = 1 as stated above).