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### Dhaka Secondary 2011/7

Posted: Fri Jan 28, 2011 10:05 pm
A six headed monster has $120$ children. He wants to give a different name to each of his children. The names will consist of $3$ English letters and one letter can be used more than once in a single name. What is the minimum number of letters the monster must use?

### Re: Dhaka Secondary 2011/7

Posted: Sat Jan 29, 2011 10:49 pm
Solution:

### Re: Dhaka Secondary 2011/7

Posted: Sat Jan 29, 2011 10:59 pm
cant we do this in this way....>
as everyname has 3 letters and repitation is allowed, so by 3 letters (a,b,c>suppose) mr. Monster can create 6 names.( $3!$ ). So he needs only 20 letters..
Anything wrong?

### Re: Dhaka Secondary 2011/7

Posted: Sun Jan 30, 2011 9:26 am
I don't think so!

### Re: Dhaka Secondary 2011/7

Posted: Sun Jan 30, 2011 11:13 am
yaap... U r right....

Actually, we need 6 letters...
$6P3=120$

### Re: Dhaka Secondary 2011/7

Posted: Tue Feb 01, 2011 9:43 am
There are many people who try the problem themselves,without knowing the solution.My solution is only to ensure they get the real answer ,and so I have to hide the solution.
The word "repetition" changes everything.It means that the name "aaa" or "bbb" is allowed. So as $5^3=125>120$
$5$ letter is enough .

### Re: Dhaka Secondary 2011/7

Posted: Tue Feb 01, 2011 11:18 am
aaaa.... 6 letters make it Comfortable ... 6 letters.. [ LoLZ ] :-p

### Re: Dhaka Secondary 2011/7

Posted: Tue Feb 01, 2011 6:48 pm
no repeatation is allowed here so the ans is $5$

### Re: Dhaka Secondary 2011/7

Posted: Tue Feb 01, 2011 8:32 pm
i just Joked man. I told if he uses 6 letters, it will be more comfortable :p

### Re: Dhaka Secondary 2011/7

Posted: Sun Jan 05, 2014 5:55 pm
*Mahi* wrote:Solution:
i got a different ans. if we follow $5^3$, we'll get some permutations twice, say BAA. but names should be different.
my solution: say $x$ is the ans. there's $x$ different names whose all of the letters are same, $^xC_2 \cdot \frac{3!}{2!}$ different names whose two letters are same and $^xP_3$ different names whose all letters are different.
$x+^xC_2 \cdot 3+^xP3$
if x=5, that'll be 95 and for 6, it'll be 171.
so the ans is 6.
let me know if i'm right.