Dhaka Secondary 2011/7

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2011/7

Unread post by BdMO » 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?

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*Mahi*
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Re: Dhaka Secondary 2011/7

Unread post by *Mahi* » Sat Jan 29, 2011 10:49 pm

Solution:
As we know that using $n$ letters $3$ times when repetition is allowed ,at most $n^3$ different words can be formed.
So the big one needs $5$ letters as $5^3=125>120$
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leonardo shawon
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Re: Dhaka Secondary 2011/7

Unread post by leonardo shawon » 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?
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*Mahi*
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Re: Dhaka Secondary 2011/7

Unread post by *Mahi* » Sun Jan 30, 2011 9:26 am

I don't think so!
Here,the name aaa or bbb is allowed.
So,it will be much less than 20
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leonardo shawon
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Re: Dhaka Secondary 2011/7

Unread post by leonardo shawon » Sun Jan 30, 2011 11:13 am

yaap... U r right....

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


please DONT Hide ur post.
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Re: Dhaka Secondary 2011/7

Unread post by *Mahi* » 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 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 .
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leonardo shawon
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Re: Dhaka Secondary 2011/7

Unread post by leonardo shawon » Tue Feb 01, 2011 11:18 am

aaaa.... 6 letters make it Comfortable ... 6 letters.. [ LoLZ ] :-p
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Tahmid Hasan
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Re: Dhaka Secondary 2011/7

Unread post by Tahmid Hasan » Tue Feb 01, 2011 6:48 pm

no repeatation is allowed here so the ans is $5$
বড় ভালবাসি তোমায়,মা

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leonardo shawon
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Re: Dhaka Secondary 2011/7

Unread post by leonardo shawon » Tue Feb 01, 2011 8:32 pm

i just Joked man. I told if he uses 6 letters, it will be more comfortable :p
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mission264
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Re: Dhaka Secondary 2011/7

Unread post by mission264 » Sun Jan 05, 2014 5:55 pm

*Mahi* wrote:Solution:
As we know that using $n$ letters $3$ times when repetition is allowed ,at most $n^3$ different words can be formed.
So the big one needs $5$ letters as $5^3=125>120$
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.

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