# Number of letters required to make three letter names

If a monster has 63 children and he wants to keep 3 letter names for each of them so that they are distinct,but with the condition that you can use the same letter more than once,how many letters at minimum does the monster need to name it's children?-source-BdMO 2011question.I have done a bit of listing and I think it is 5. As others pointed out,my listing was wrong.

Let $n$ be the number of letters we need to form $63$ distinct three-letter names.

So, we have $n$ choices for the first letter, $n$ choices for the second letter and $n$ choices for the third letter: This means we have $n\times n\times n = n^3$ possible combinations of distinct names.

We need $n^3 \geq 63$ where $n$ is an integer. $n = 4$ gives us $4^3 = 64$ distinct names, and this is indeed the least integer giving us at least $63$ distinct names. (For $n = 3$, we have only $3^3 = 27$ possible distinct names, and for $n = 5$, we have as many as $5^3 = 125$ distinct three letter names.

Hence, the monster needs minimally $4$ distinct letters to create 63 distinct three-letter names.

• Good ans,you explained it clearly. – user83718 Jun 24 '13 at 17:28
• You're welcome! ;-) – amWhy Jun 24 '13 at 17:30
• Now I get why the original question mentioned the monster has 4 heads. – user83718 Jun 24 '13 at 17:36
• @amWhy: So nice to get that sort of feedback! :-) TU +1 – Amzoti Jun 25 '13 at 0:13
• This is not an intersection but a Horseshoe for being lucky for ever. $\LARGE{\bf{\cap}}$ – mrs Jun 30 '13 at 4:38

Hints:

• Does the choice of a first letter influence the second and third letters?
• $4^3 = 64$.
• NB. I interpreted your question as allowing the same letter to occur multiple times in a name. – Lord_Farin Jun 24 '13 at 17:16
• Yes,You interpreted it correctly – user83718 Jun 24 '13 at 17:19

On the assumption that order matters, so you can use both aab and aba, for $n$ letters there are $n^3$ names and we need $n=4$.

If order does not matter, we have $n$ names with all letters the same, $n(n-1)$ with two of one letter and one of another, and $\frac 16n(n-1)n-2$ with all letters different. This adds up to $\frac 16n^3+\frac 12n^2+\frac 13n$ and we need $n=7$

• In case of names,order does matter. – user83718 Jun 24 '13 at 17:23
• @user83718: I thought it likely, especially as the number of children is closer to the limit in that case. But I didn't know if the monster can keep track of different orders, so I did it both ways. – Ross Millikan Jun 24 '13 at 17:25
• Haha,appreciate your effort though. – user83718 Jun 24 '13 at 17:26

The answers listed above have attached an unspecified paradigm - i.e.visual form of the names. The Question states that this is a monster, not a man, and the criteria is "distinction". Why do we seek an answer in the written form? Is this a known characteristic of monsters?

Webster - DISTINCTION : a difference that you can see, hear, smell, feel, etc. : a noticeable difference between things or people

Wouldn't 3 letters, pronounced at different octaves, or in ascending tones, or descending tones, etc. meet the test of distinction? and since letters may be repeated..........