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?
BdMO National Higher Secondary 2009/6
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: BdMO National Higher Secondary 2009/6
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,
$38+38\cdot 37+36=38^2+36=1480.$
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
My answer is $1520$ ... I am a novice in combi... Maybe I got this wrong...
And Labib I didn't understand this...
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...
- bristy1588
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Re: BdMO National Higher Secondary 2009/6
@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.
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
Re: BdMO National Higher Secondary 2009/6
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
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
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
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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Re: BdMO National Higher Secondary 2009/6
The correct answer is 1480.