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### Dhaka Secondary 2011/7

Posted: **Fri Jan 28, 2011 10:05 pm**

by **BdMO**

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**

by ***Mahi***

### Re: Dhaka Secondary 2011/7

Posted: **Sat Jan 29, 2011 10:59 pm**

by **leonardo shawon**

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**

by ***Mahi***

### Re: Dhaka Secondary 2011/7

Posted: **Sun Jan 30, 2011 11:13 am**

by **leonardo shawon**

yaap... U r right....

Actually, we need 6 letters...

$6P3=120$

please DONT Hide ur post.

### Re: Dhaka Secondary 2011/7

Posted: **Tue Feb 01, 2011 9:43 am**

by ***Mahi***

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 answer is NOT 6.

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**

by **leonardo shawon**

aaaa.... 6 letters make it Comfortable ... 6 letters.. [ LoLZ ] :-p

### Re: Dhaka Secondary 2011/7

Posted: **Tue Feb 01, 2011 6:48 pm**

by **Tahmid Hasan**

no repeatation is allowed here so the ans is $5$

### Re: Dhaka Secondary 2011/7

Posted: **Tue Feb 01, 2011 8:32 pm**

by **leonardo shawon**

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**

by **mission264**

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.