Help me

For discussing Olympiad Level Combinatorics problems
atiqur_jhe
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Help me

Unread post by atiqur_jhe » Sun Apr 17, 2011 8:07 pm

Let X=(1,2,3,,,,100) and let S=((a,b,c):a,b,c€X,a<b and a<c) find the number of S. there is some prob in 2nd bracket so i cant give it.

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*Mahi*
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Re: Help me

Unread post by *Mahi* » Thu Apr 21, 2011 1:10 pm

There can be $100$ different choice for $a$ , and $100-a$ for $b$ & $c$ each ,so the answer is
$\sum_{n=1}^{100} n^2$ or $338350$ .
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rakeen
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Re: Help me

Unread post by rakeen » Sat May 21, 2011 7:56 pm

@mahi I think atiqur wrote $a<b$ LESS THAN. And even though if you thought it is $a>b$ then how did you figured out the sum?
I mean why it is: $\sum_{n=1}^{100} n^2$

Suppose, $X$ $=$ {$1$,$2$,$3$}
If $a=1$ , then the number of $S$ is $4$. S={1,2,2}, {1,2,3}, {1,3,2}
And if $a=2$ , then the number of $S$ is $1$ S={2,3,3}
So, the total number of $S$ is $4$ in this case.
But according to your method it should be: $\sum_{n=1}^{3} n^2$ or, $14$ .

@atiqur_jhe well, $2nd$ bracket is just beside $P$. And if you're talking about $L_A T_E X$ then you might see this: matholympiad.org.bd/forum/viewtopic.php@f=25&t=2.htm (How to use LATEX)
It'll take 10 minutes to learn some basic tatice of latex for general use.
r@k€€/|/

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*Mahi*
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Re: Help me

Unread post by *Mahi* » Fri May 27, 2011 11:49 am

For every choice of $a$ , there are $100-a$ elements greater than $a$ in the set , all of which is a choice for $b,c$ .
Please read Forum Guide and Rules before you post.

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

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rakeen
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Re: Help me

Unread post by rakeen » Mon May 30, 2011 9:47 am

yes. so...........
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