Dhaka Secondary 2011/7
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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.
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.
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
Solution:
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Nur Muhammad Shafiullah | Mahi
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- leonardo shawon
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Re: Dhaka Secondary 2011/7
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?
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 .....
BRAC University.
long way to go .....
Re: Dhaka Secondary 2011/7
I don't think so!
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Nur Muhammad Shafiullah | Mahi
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Nur Muhammad Shafiullah | Mahi
- leonardo shawon
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Re: Dhaka Secondary 2011/7
yaap... U r right....
Actually, we need 6 letters...
$6P3=120$
please DONT Hide ur post.
Actually, we need 6 letters...
$6P3=120$
please DONT Hide ur post.
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Secondary 2011/7
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 .
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
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- leonardo shawon
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- Location:Dhaka
Re: Dhaka Secondary 2011/7
aaaa.... 6 letters make it Comfortable ... 6 letters.. [ LoLZ ] :-p
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
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- Tahmid Hasan
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- leonardo shawon
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Re: Dhaka Secondary 2011/7
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 .....
BRAC University.
long way to go .....
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Re: Dhaka Secondary 2011/7
i got a different ans. if we follow $5^3$, we'll get some permutations twice, say BAA. but names should be different.*Mahi* wrote:Solution:
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.