BdMO National Higher Secondary 2009/6

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Moon
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BdMO National Higher Secondary 2009/6

Unread post by Moon » Sun Feb 06, 2011 11:33 pm

Problem 6:
Forty MOVers (Mathematical Olympiad Volunteers) are sitting in a circle. Munir Hasan randomly chooses $3$ volunteers to help in the awards ceremony. In how many ways can the volunteers be chosen such that at least $2$ of the volunteers were sitting next to each before being chosen?
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Labib
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Re: BdMO National Higher Secondary 2009/6

Unread post by Labib » Sun Dec 25, 2011 4:32 am

Here's how I thought it should be done...
There are 40 possible pairs...
choosing any one pair, I can choose the 3rd MOVer in 38 ways.
The next pair will have 37 choices and it will stay like this until the last pair steps in having 36 ways to choose the 3rd MOVer.
Thus total way,
$38+38\cdot 37+36=38^2+36=1480.$
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Abdul Muntakim Rafi
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Re: BdMO National Higher Secondary 2009/6

Unread post by Abdul Muntakim Rafi » Sun Dec 25, 2011 10:45 pm

My answer is $1520$ ... I am a novice in combi... Maybe I got this wrong... :?:
And Labib I didn't understand this...
The next pair will have 37 choices and it will stay like this until the last pair steps in having 36 ways to choose the 3rd MOVer.
Thus total way,
38+38*37+36=382+36=1480:
Man himself is the master of his fate...

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bristy1588
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Re: BdMO National Higher Secondary 2009/6

Unread post by bristy1588 » Sat Jan 14, 2012 2:00 pm

@Labib Rashid:
Could you explain your solution again?
@Rafi, How did u get 1520??

Well,maybe I am wrong but here is my solution:
From 40 movers, 3 movers can be chosen in $\binom{40}{3}$ ways. Now, let us count the number of ways that 3 movers can be selected such that there are no movers who sat next to each other. This can be done in $2*\binom{20}{3}$. Therefore, in the other combinations there must be atleast 2 movers who sat next to each other.
So, total number of Ways:=$\binom{40}{3}-2*\binom{20}{3}=7600$ ways,

If my answer is wrong, please explain where i am wrong.
Bristy Sikder

Shapnil
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Re: BdMO National Higher Secondary 2009/6

Unread post by Shapnil » Thu Mar 22, 2012 8:43 pm

my answer is also 1480.i may be wrong bt i'm trying to prove this
first we consider that we'll chose three movers where all of them r sitting next to each other.we can select them in 40 ways.
secondly,we can chose 2 movers in 40 ways.w\so, we've to chose 1 movers from (40-4) or 36 movers.
then total number of ways :40+(40*36) =1480
is it correct :?: :?: :?:

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Re: BdMO National Higher Secondary 2009/6

Unread post by shehab ahmed » Thu Mar 22, 2012 11:13 pm

দুটি জিনিস ঘটতে পারে।
১.যেই তিনজনকে নেয়া হবে তারা সকলেই পাশাপাশি বসা ছিল।এটি করা যায় ৪০ ভাবে।
২.যেই তিনজনকে নেয়া হয়েছিল তাদের শুধুমাত্র দুইজন পাশাপাশি বসা ছিল।এই দুইজনকে নেয়া যায় ৪০ ভাবে।তারপর অপর একজনকে নেয়া যাবে ৩৬ ভাবে।তাই এটি করা যায় ৪০*৩৬ ভাবে।
তাই উত্তর হল ৪০*৩৭

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samiul_samin
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Re: BdMO National Higher Secondary 2009/6

Unread post by samiul_samin » Sat Dec 30, 2017 9:23 pm

The correct answer is 1480.

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