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Question:

Given the letters $a, b, c.$ How many ways to arrange these letters?

Assuming 3 letters.

Logical Question:

How do you know if you need to use permutations or combinations for this question ?

$P(3,3)$ or $C(3,3)$ ?

2 Answers 2

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You need to use permutations, so the answer is $3!=6$. Using combination does not make much sense since it would be equivalent to count the number of ways you can take $3$ letters out of a set of $3$ letters, which is only $1$. Indeed $C(3,3)=\frac{3!}{1!\cdot 3!}=1$.

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    Ok I see what you mean, I only use combinations if I am ask to do so, else I need to use permutations. – 2017-02-17
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    You use combination whenever the order does not count. If the order wasn't important in your question, then $abc$ and $bca$ (for instance) would be considered the same. – 2017-02-17
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    I see so abc and bca would have been considered 1 arrangement which is $3\choose3$ = 1 – 2017-02-17
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    Thanks for your clarification. – 2017-02-17
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    Yes, if the order didn't count all the $6$ combinations $abc,acb,bac,bca,cab,cba$ would be considered exactly the same. – 2017-02-17
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The order of the letters is important here. Hence you need to use permutations.

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    Why is it important ? You can start with either of a, b, c. – 2017-02-17
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    Well, you just answered your own question!! How do you count the number of arrangements of the letters $a,b,c$? First you need to choose the first letter: $3$ choices. Then you choose the second letter from the remaining $2$: $2$ choices. Then you have $1$ choice for the last letter, the letter that wasn't chosen already. By the rule of product, the number you seek is $3\cdot2\cdot1=3!$ which is the number of permutations of the three letters. – 2017-02-17
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    The number of $3$-combinations of the $3$ letters will give you $1$. In a sense, it gives you the number of arrangements in alphabetical order only and so it counts only the arrangement $abc$. – 2017-02-17
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    So you mean if I am asked to find the # of arrangements of a,b,c I use permutations or 3! unless the question explicitly asks to find the # of combinations of a,b,c ? – 2017-02-17
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    Basically, yeah. – 2017-02-17
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    Ok makes sense. Thank you for your help. – 2017-02-17
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    so simply there are variations without repetitions – 2017-02-18