Dhaka Secondary 2011/7

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
Posts:134
Joined:Tue Jan 18, 2011 1:31 pm
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?

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

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 $5^3=125>120$
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
leonardo shawon
Posts:169
Joined:Sat Jan 01, 2011 4:59 pm
Location:Dhaka

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?
Ibtehaz Shawon
BRAC University.

long way to go .....

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

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
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
leonardo shawon
Posts:169
Joined:Sat Jan 01, 2011 4:59 pm
Location:Dhaka

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.
Ibtehaz Shawon
BRAC University.

long way to go .....

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

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 .
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
leonardo shawon
Posts:169
Joined:Sat Jan 01, 2011 4:59 pm
Location:Dhaka

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
Ibtehaz Shawon
BRAC University.

long way to go .....

User avatar
Tahmid Hasan
Posts:665
Joined:Thu Dec 09, 2010 5:34 pm
Location:Khulna,Bangladesh.

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

User avatar
leonardo shawon
Posts:169
Joined:Sat Jan 01, 2011 4:59 pm
Location:Dhaka

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
Ibtehaz Shawon
BRAC University.

long way to go .....

mission264
Posts:6
Joined:Sun Jan 05, 2014 5:27 pm

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.

Post Reply