BdMO National 2012: Higher Secondary 02
Problem 2:
Superman is taking part in a hurdle race with $12$ hurdles. At any stage he can jump across any number of hurdles lying ahead. For example, he can cross all $12$ hurdles in one jump or he can cross $7$ hurdles in the first jump, $1$ in the later and the rest in the third jump. In how many different ways can superman complete the race?
Superman is taking part in a hurdle race with $12$ hurdles. At any stage he can jump across any number of hurdles lying ahead. For example, he can cross all $12$ hurdles in one jump or he can cross $7$ hurdles in the first jump, $1$ in the later and the rest in the third jump. In how many different ways can superman complete the race?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: BdMO National 2012: Higher Secondary 02
আমার ২ টা সমাধান আছে কিন্তু যতদুর ধারনা ১ম টা রাইট। কিন্তু ২য় টা কেন ভুল সেটা জানা দরকার।
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
- zadid xcalibured
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Re: BdMO National 2012: Higher Secondary 02
11 gaps between 12 hurdles.so each way corresponds to a choice of this gaps.so the answer is 2^11
Re: BdMO National 2012: Higher Secondary 02
My question is that, where is my mistake ???zadid xcalibured wrote:11 gaps between 12 hurdles.so each way corresponds to a choice of this gaps.so the answer is 2^11
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
- zadid xcalibured
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- Location:mymensingh
Re: BdMO National 2012: Higher Secondary 02
i see boxes in stead of ur solution.
Re: BdMO National 2012: Higher Secondary 02
লাল অংশটা এবং সাথে এর পরে যা যা ধরস (৪টা ৪ ভাবে, ৫টা ৫ ভাবে... ) এইগুলা ভুল হইসে।sm.joty wrote: ....
অর্থাৎ ১ম এ একবারে ১১ টা পার হবে তারপর ১ টা। ১ ভাবে।
২য় ক্ষেত্রে প্রথম ১০ টা একবারে তারপর বাকি ২ টা যাওয়া যায় ২ ভাবে।
৩য় ক্ষেত্রে প্রথম ৯ টা একবারে তারপর বাকি ৩ টা যাওয়া যায় ৩ ভাবে।
....
Every logical solution to a problem has its own beauty.
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- nafistiham
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Re: BdMO National 2012: Higher Secondary 02
well,this one can be done like this,too,i think.
$11$ gaps.
so, the choices will be
\[\sum_{k=1}^{11}\binom{11}{k}=2^{11}\]
$11$ gaps.
so, the choices will be
\[\sum_{k=1}^{11}\binom{11}{k}=2^{11}\]
Last edited by nafistiham on Thu Feb 16, 2012 7:17 pm, edited 1 time in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Cryptic.shohag
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Re: BdMO National 2012: Higher Secondary 02
Number of ways when he doesn't land is 1.
Number of ways when he lands once is 11C0.
Number of ways when he lands twice is 11C1.
Number of ways when he lands thrice is 11C2.
Thus for 4,5,.....,11 landings the number of ways would be 11C3, 11C4,......, 11C10 respectively.
So, in total the number of ways to complete the race is 1+11C0+11C1+........+11C10= 2048.
Number of ways when he lands once is 11C0.
Number of ways when he lands twice is 11C1.
Number of ways when he lands thrice is 11C2.
Thus for 4,5,.....,11 landings the number of ways would be 11C3, 11C4,......, 11C10 respectively.
So, in total the number of ways to complete the race is 1+11C0+11C1+........+11C10= 2048.
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein
- Tahmid Hasan
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Re: BdMO National 2012: Higher Secondary 02
there are $11$ spots between $12$ hurdles,you can either land or 'not land' on those hurdles;so we have $2$ possible choices for each spot.so the answer is $2^{11}$
বড় ভালবাসি তোমায়,মা
Re: BdMO National 2012: Higher Secondary 02
using bionomial coefficient it's ans 2 to the power 11