What are the difference between Combination and Permutations?

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kundanad
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I can't able to figure out which questions are combination and permutations.
Even though I can figure out by the question, How can I solve it?
I am confused. Please make it more clear.
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K-Man_PhysCheM
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(Original post by kundanad)
I can't able to figure out which questions are combination and permutations.
Even though I can figure out by the question, How can I solve it?
I am confused. Please make it more clear.
Both are for finding the number of different arrangements of objects.

With permutations, the order in which the objects are arranged matters. So If you need to pick 2 from five objects A, B, C, D, E, the possible permutations would be:
AB; AC; AD; AE; BA; BC; BD; BE; CA; CB; CD; CE; DA; DB; DC; DE; EA; EB; EC; ED

The first slot can be any of the five objects, and the second slot can be any of the four remaining objects. Order doesn't matter, so AB is different to BA. Therefore the total number of permutations is 5\times 4 = 20.

For n objects where you pick r out, there are \dfrac{n!}{(n-r)!} permutations, where "!" is factorial. 5! \equiv 5\times 4\times 3\times 2\times 1

In our example above, n=5 and r=2, so (n-r) = (5-2) = 3, and so the total number of permutations is \dfrac{5!}{3!} = \dfrac{5\times 4\times 3!}{3!} = 5 \times 4 = 20, as before.

You can calculate permutations quickly with your calculator by typing in n\mathrm{P}r, where n is the total number of objects and r is the number you are picking. In this example, you could type in 5\mathrm{P}2, which would give 20.

You tend to use permutations when working out possible numerical codes, where the order in which you input the numbers does matter. EG: How many different 5-digit codes can I make with the digits 0 to 9 without repetition?

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With combinations, the order in which the objects are arranged does NOT matter. So AB is equivalent to BA, so would only be counted once.

There will always be a factor of r repeats, so the number of combinations of arranging n objects in r slots is \dfrac{n!}{r!(n-r)!}

You can calculate these quickly using the nCr button on your calculator, as with permutations.

You tend to use combinations when picking a group of people where the order in which they are picked doesn't change anything. EG Alice, Bob, Charlie, David and Emily apply for the same job. There are three equivalent job spaces available. In how many ways can they be chosen?

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How to go about answering permutations and combinations problems:

The problem tends to lie in which one to choose. Intuition will come with practice, but generally use combinations when you see the word "equivalent" or similar, or when the equivalence is implied by the situation, eg getting free tickets.

Use permutations when the order matters. Eg, if 10 people are auditioning for three different characters in a play, the number of possible arrangements is given by permutations, since the order DOES matter as the characters are different.

Often it may be helpful to draw a diagram of slots/available positions and objects. Struggle through the questions without looking at the answers. Practice a lot of questions, and you will find variations of the same question repeat.

For more specific help, it would be worth posting a few questions that you are struggling with so that we may give you some hints.
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kundanad
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Thank you for your help.
I understood bit and I am sure it will be more clear from these questions.

A group of 9 people consists of 2 boys, 3 girls and 4 adults. In how many ways can a team of 4 be chosen if
(i) both boys are in the team
(ii) the adults are either all in the team or all not in the team
(iii) at least 2 girls are in the team?

The back row of a cinema has 12 seats, all of which are empty. A group of 8 people, including Mary and Frances, sit in this row.
Find the number of different ways they can sit in these 12 seats if
(i) there are no restrictions
(ii) Mary and Frances do not sit in seats which are next to each other
(iii) all 8 people sit together with no empty seats between them



(Original post by K-Man_PhysCheM)
Both are for finding the number of different arrangements of objects.

With permutations, the order in which the objects are arranged matters. So If you need to pick 2 from five objects A, B, C, D, E, the possible permutations would be:
AB; AC; AD; AE; BA; BC; BD; BE; CA; CB; CD; CE; DA; DB; DC; DE; EA; EB; EC; ED

The first slot can be any of the five objects, and the second slot can be any of the four remaining objects. Order doesn't matter, so AB is different to BA. Therefore the total number of permutations is 5\times 4 = 20.

For n objects where you pick r out, there are \dfrac{n!}{(n-r)!} permutations, where "!" is factorial. 5! \equiv 5\times 4\times 3\times 2\times 1

In our example above, n=5 and r=2, so (n-r) = (5-2) = 3, and so the total number of permutations is \dfrac{5!}{3!} = \dfrac{5\times 4\times 3!}{3!} = 5 \times 4 = 20, as before.

You can calculate permutations quickly with your calculator by typing in n\mathrm{P}r, where n is the total number of objects and r is the number you are picking. In this example, you could type in 5\mathrm{P}2, which would give 20.

You tend to use permutations when working out possible numerical codes, where the order in which you input the numbers does matter. EG: How many different 5-digit codes can I make with the digits 0 to 9 without repetition?

-----
With combinations, the order in which the objects are arranged does NOT matter. So AB is equivalent to BA, so would only be counted once.

There will always be a factor of r repeats, so the number of combinations of arranging n objects in r slots is \dfrac{n!}{r!(n-r)!}

You can calculate these quickly using the nCr button on your calculator, as with permutations.

You tend to use combinations when picking a group of people where the order in which they are picked doesn't change anything. EG Alice, Bob, Charlie, David and Emily apply for the same job. There are three equivalent job spaces available. In how many ways can they be chosen?

----
How to go about answering permutations and combinations problems:

The problem tends to lie in which one to choose. Intuition will come with practice, but generally use combinations when you see the word "equivalent" or similar, or when the equivalence is implied by the situation, eg getting free tickets.

Use permutations when the order matters. Eg, if 10 people are auditioning for three different characters in a play, the number of possible arrangements is given by permutations, since the order DOES matter as the characters are different.

Often it may be helpful to draw a diagram of slots/available positions and objects. Struggle through the questions without looking at the answers. Practice a lot of questions, and you will find variations of the same question repeat.

For more specific help, it would be worth posting a few questions that you are struggling with so that we may give you some hints.
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K-Man_PhysCheM
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(Original post by kundanad)
Thank you for your help.
I understood bit and I am sure it will be more clear from these questions.

A group of 9 people consists of 2 boys, 3 girls and 4 adults. In how many ways can a team of 4 be chosen if
(i) both boys are in the team
(ii) the adults are either all in the team or all not in the team
(iii) at least 2 girls are in the team?

The back row of a cinema has 12 seats, all of which are empty. A group of 8 people, including Mary and Frances, sit in this row.
Find the number of different ways they can sit in these 12 seats if
(i) there are no restrictions
(ii) Mary and Frances do not sit in seats which are next to each other
(iii) all 8 people sit together with no empty seats between them
The first one is about choosing a team, so you want to use combinations.

For part (i), both boys HAVE to be in the team. Therefore you are only choosing two more people for the team from the 7 remaining people.

Part (ii), split it into both cases, where all adults are in team and no adults are in team, and add up both combinations.

Part (iii), split it into two cases: one with two girls and one with three girls. The three girls case should be easy, since you can just pick one person from the remaining 6. The two girl case is a little trickier, but note you are choosing two girls from three and then two others from 6 remaining non-girls. Add these up.

Question 2 can be approached similarly.

Please post your working .though process too so we can give more targeted help
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Physics Enemy
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(Original post by kundanad)
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Combinations are unique selections of r objects from a total of n. Ordering/arrangement doesn't matter. Selecting 1&2 (or 2&1) counts as one, 2-object combination from e.g) numbers 1 to 5.

Permutations extends this concept to include reordering/rearranging of said combinations, so we get far more of them. So 1&2 and 2&1 are two, 2-object permutations from e.g) numbers 1 to 5.
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bibekpandey
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Let me illustrate with the help of an example.

case i) how many different numbers can be made from the digits 1,2 and 3?
ii)how many different subsets can be made from 1,2 and 3?
solution:
i) here you need to form a number. say you formed 321 now zig-zag the numbers and you get 123 or 132.
are these numbers same??? Absolutely no!! here we use permutation
ii) Here you need to form a subset . say you took one of them {1,2,3} .Now zigzag the numbers , and you will get {3,2,1}. Are they same??? yes! Now you use combination in this case..

This is the best and easiest way of finding whether the question uses permutation or combination... Thanks!!!
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kadammanali987
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Combination: Combination is simply referred to the selection process.
Lets us now elaborate it with the help of an example. Suppose we have three students named as A,B,C in a class and the teacher want to select two students for a quiz. The all possible ways in which this process can be done are given below:-

A, B
A, C
B, C
Here A,B and B,A are one and the same as in combination we do not bother about the ordering of the objects i.e. arrangement does not matters.

The mathematical formula combination is:

C(n, r) = n!/ ((n-r)!*r!)

Here C(n, r) refers to the number of ways of selecting r objects from a group of n objects.

2) Permutation: Permutation is simply referred to the arrangement process.

Let us now understand this with the help of an example. Suppose we have three objects named A, B, C and we want to arrange them taken 2 at a time. The all possible arrangements are given below:-

A, B
B, A
A, C
C, A
B, C
C, B
Here the order of the objects does matters. For example A, B and B, A are two different arrangements. Hence there is a importance of order of the arrangements.

Mathematical formula for Permutation is:-

P(n, r) = n!/(n-r)!

P(n, r) refers to the number of arrangements of n objects taken r at a time.
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DFranklin
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Not much point replying to a thread a year old guys...
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