## BdMO National Higher Secondary 2009/6

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
Moon
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Contact:

### BdMO National Higher Secondary 2009/6

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?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### 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,
$38+38\cdot 37+36=38^2+36=1480.$
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Abdul Muntakim Rafi
Posts: 173
Joined: Tue Mar 29, 2011 10:07 pm
Location: bangladesh,the earth,milkyway,local group.

### 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...
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
Posts: 92
Joined: Sun Jun 19, 2011 10:31 am

### 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.
Bristy Sikder

Shapnil
Posts: 10
Joined: Mon Dec 19, 2011 2:05 pm

### 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

shehab ahmed
Posts: 66
Joined: Tue Mar 20, 2012 12:52 am

### 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

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National Higher Secondary 2009/6

The correct answer is 1480.