Try to give me an example and steps on how to do Permutations and Combinations WITHOUT A CALCULATOR.
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Let's start with permutations.
EXAMPLE 1
How many ways can you permute A B C D?
Well, you have 4 letters, so you have a choice of 4 for the first place, and that leaves 3 for the next, then 2 then 1. So the total number of permutations is 4x3x2x1 = 24. (You have to be able to multiply and divide without a calculator too, but let's presume that won't be a problem.)
The formula for this is:
Permutation of n things taken n at at time: n!
EXAMPLE 2
Now suppose you have 6 letters, A B C D E F, and you want to take just 2 at a time. You have 6 choices for the first place, then 5, then you stop. So it's 6x5 = 30.
Note that this is also 6! / 4!. The 4! in the denominator cancels out the 4x3x2x1 part of 6!, leaving you only 6x5.
The formula for this is:
Permutation of n things taken k at a time: n! / (k-1)!.
COMBINATIONS
Combinations are done mostly the same way, but you have to allow for duplicates in the permutations. E.g., in the first example above,
ABCD
BCDA
DABC
and others are the same combination of letters.
You cancel out the duplicates by dividing by k!, where k is the number items you are taking each time. For example, in EXAMPLE 2, how many duplicates do you have for each permutation, where you are taking 2 things at at time?
You have AB and BA, and that's it.
You have AC and CA, AD and DA, etc. So you can see that you will have 2 permuations for each combination.
What if you had 3 at a time?
You'd have ABC ACB BAC BCA CAB CBA, or 6 repetitions.
In general, when you are taking k things at a time, you have to divide the permutations by k! to allow for the repetitions.
The general formula is:
Combination of n things taken k at a time is: n! / ( (k-1)! k! )
So in the second example, where you have 5 things taken 2 at a time, you have:
6x5x4x3x2x1
divided by
4x3x2x1 times 2x1
= 6x5 / 2x1 = 15.
You can check that:
AB AC AD AE AF
BC BD BE BF
CD CE CF
DE DF
EF
I hope this helps!
EXAMPLE 1
How many ways can you permute A B C D?
Well, you have 4 letters, so you have a choice of 4 for the first place, and that leaves 3 for the next, then 2 then 1. So the total number of permutations is 4x3x2x1 = 24. (You have to be able to multiply and divide without a calculator too, but let's presume that won't be a problem.)
The formula for this is:
Permutation of n things taken n at at time: n!
EXAMPLE 2
Now suppose you have 6 letters, A B C D E F, and you want to take just 2 at a time. You have 6 choices for the first place, then 5, then you stop. So it's 6x5 = 30.
Note that this is also 6! / 4!. The 4! in the denominator cancels out the 4x3x2x1 part of 6!, leaving you only 6x5.
The formula for this is:
Permutation of n things taken k at a time: n! / (k-1)!.
COMBINATIONS
Combinations are done mostly the same way, but you have to allow for duplicates in the permutations. E.g., in the first example above,
ABCD
BCDA
DABC
and others are the same combination of letters.
You cancel out the duplicates by dividing by k!, where k is the number items you are taking each time. For example, in EXAMPLE 2, how many duplicates do you have for each permutation, where you are taking 2 things at at time?
You have AB and BA, and that's it.
You have AC and CA, AD and DA, etc. So you can see that you will have 2 permuations for each combination.
What if you had 3 at a time?
You'd have ABC ACB BAC BCA CAB CBA, or 6 repetitions.
In general, when you are taking k things at a time, you have to divide the permutations by k! to allow for the repetitions.
The general formula is:
Combination of n things taken k at a time is: n! / ( (k-1)! k! )
So in the second example, where you have 5 things taken 2 at a time, you have:
6x5x4x3x2x1
divided by
4x3x2x1 times 2x1
= 6x5 / 2x1 = 15.
You can check that:
AB AC AD AE AF
BC BD BE BF
CD CE CF
DE DF
EF
I hope this helps!
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P(n,r) = n!/(n-r)!
C(n,r) = P(n,r)/r! = n!/(n-r)!(r)!
So, for example, say you have 6 books and your wanted to find the number of ways to order 4 of them. The answer would be P(6,4) = 6!/2! = 6*5*4*3 = 360.
Now, say you just wanted to choose 4 out of the 6 books. The answer would be C(6,4) = P(6,4)/4! = 360/24 = 15, or C(6,4) = 6!/(4!2!) = 720/48 = 15.
C(n,r) = P(n,r)/r! = n!/(n-r)!(r)!
So, for example, say you have 6 books and your wanted to find the number of ways to order 4 of them. The answer would be P(6,4) = 6!/2! = 6*5*4*3 = 360.
Now, say you just wanted to choose 4 out of the 6 books. The answer would be C(6,4) = P(6,4)/4! = 360/24 = 15, or C(6,4) = 6!/(4!2!) = 720/48 = 15.