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I am having trouble distinguishing when to use factorials, $^nC_r$ and $^nP_r$.

I am considering this description of their various purposes:

"Factorials serve two purposes, both stated above: the number of ways to order a set of $n$ objects is $n! $, and this makes it useful in other formulas. What you call "$^nC_r$" stands for combinations - the number of ways to choose $r$ objects from a set of $n$ distinguishable objects, where order doesn't matter. Contrast this with $^nP_r$, which is the number of ways to choose $r$ objects from a set of $n$ distinguishable objects when order does matter."

I don't really understand what is meant by 'order matters.'

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    Order matters means that $abc$ and $bca$ are different selections. – 2017-02-14
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    @agra94 Please note the tag description for probability-theory: "Use this tag only if your question is about the modern theoretical footing for probability, for example probability spaces, random variables, law of large numbers, and central limit theorems." – 2017-02-14

4 Answers 4

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C is for combination, and P for permutation. Think of C as a group and P as a list.

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Chosing $r$ items from a set of $n$ can be done in

  • $^nP_r=n(n-1)(n-2)\cdots(n-r+1)=\dfrac{n!}{(n-r)!}$ ways when order matters, because you choose among $n$, then among $n-1$, then $n-2$, ... down to $n-r+1$.

  • $^nC_r=\dfrac{n!}{r!(n-r)!}=\dfrac{^nP_r}{r!}$ ways when order doesn't matter, because every selection appears exactly $r!$ times, by permutation.

Two limit cases can be considered.

  • when $r=0$, you can choose no element in a single way, and $^nP_r=^nC_r=1$.

  • when $r=n$, you choose all elements in $^nP_r=n!$ ways when order matters, and in a single way, $^nC_n=1$ otherwise, as you simply pick them all.

Examples:

$^3P_2=6:ab,ac,ba,bc,ca,cb$

$^3C_2=\dfrac62=3:ab,ac,bc$

$^3P_3=6:abc,acb,bac,bca,cab,cba$

$^3C_3=\dfrac66=1:abc$

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    @learnandshare: if you change the notation, change it in the OP as well. – 2017-02-15
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Imagine you have $5$ objects in the set $S=\{1,2,3,4,5\}$.

A permutation defines an ordering in the whole set of $5$ elements:

$\{1,2,5,4\},\{1,4,2,5\},\{1,4,5,2\}...$ and the total orderings is $5!$

A combination for example $5C3$ chooses $3$ objects from the total of $5$ objects:

$\{1,2,3\},\{1,2,4\},\{1,2,5\}...$ but since order does not matter, $\{1,2,3\}$ and $\{3,2,1\}$ is the same group and its formula is $\frac{n!}{r!(n-r)!}$

If order matters, then $\{1,2,3\}$ and $\{3,2,1\}$ are different, this is called permutation of a combination and its formula is $\frac{n!}{(n-r)!}$. Note that you are permuting each combination, that is why the $r!$ factorial of the denominator is canceled.

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I will just give a short explanation on that. For combinations the $\textbf{orders do not matter}$. For example there are $4$ person we need to choose $2$ of them so we use $^4C_2$ to obtain $6$.

Let the $4$ person be $p_1,p_2,p_3,p_4$, we have choosen $2$ from $4$ so the combinations are as follows:

$(p_1,p_2),$$(p_3,p_4),$$(p_2,p_3),$$(p_2,p_4),$$(p_1,p_4),$$(p_1,p_3)$

Now, let's move on to permutation. Suppose we need to permutate $2$ out of the $4$ person, mathematically it is written as $^4P_2$ to obtain $12$. In this case $\textbf{order matters}$, the permuations are as follows:

$(p_1,p_2),$$(p_3,p_4),$$(p_2,p_3),$$(p_2,p_4),$$(p_1,p_4),$$(p_1,p_3)$,$(p_2,p_1),$$(p_4,p_3),$$(p_3,p_2),$$(p_4,p_2),$$(p_4,p_1),$$(p_3,p_1)$