Can any one find out the mistake in it?
BdMO National 2012: Higher Secondary 02
Re: BdMO National 2012: Higher Secondary 02
In general we can assume that any natural number '$n$' can be shown as the sum of '$k$' positive integers in $(2^n-1)-1$ ways where $0<k<n$.
so the solution is $2^{12-1}-1+1=2^{11}$
so the solution is $2^{12-1}-1+1=2^{11}$
Last edited by Phlembac Adib Hasan on Sun Feb 10, 2013 11:29 am, edited 2 times in total.
Reason: $L^AT_EX$-ed
Reason: $L^AT_EX$-ed
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Re: BdMO National 2012: Higher Secondary 02
Let's make a pattern then solve this..
Suppose, we've 1 hurdle, so he can jump only in 1 way.
If we'd 2 hurdles, he could jump in 2 different ways.
If there is 3 hurdles, he can jump in 4 different ways.
If there is 4 hurdles, he can jump in 7 different ways.
If there is 5 hurdles, he can jump in 11 different ways.
Now we've 1, 2, 4, 7, 11
\/ \/ \/ \/
1 2 3 4
\/ \/ \/
1 1 1
Now we can say 11=(4+3+2+1)+1
There were 5 hurdles to get 11 different ways..but we've 4 terms in the pattern...let the 4=n and 5=H.. so n=H-1.. The number of hurdles is H.
So, 11=4(4+1)/2 + 1
C=n(n+1)/2 + 1 [The number of ways is C]
C=(H-1)H/2 + 1
Now we put the value of H=12
We'll get C=67
So,There are 67 different ways to complete the race.
Suppose, we've 1 hurdle, so he can jump only in 1 way.
If we'd 2 hurdles, he could jump in 2 different ways.
If there is 3 hurdles, he can jump in 4 different ways.
If there is 4 hurdles, he can jump in 7 different ways.
If there is 5 hurdles, he can jump in 11 different ways.
Now we've 1, 2, 4, 7, 11
\/ \/ \/ \/
1 2 3 4
\/ \/ \/
1 1 1
Now we can say 11=(4+3+2+1)+1
There were 5 hurdles to get 11 different ways..but we've 4 terms in the pattern...let the 4=n and 5=H.. so n=H-1.. The number of hurdles is H.
So, 11=4(4+1)/2 + 1
C=n(n+1)/2 + 1 [The number of ways is C]
C=(H-1)H/2 + 1
Now we put the value of H=12
We'll get C=67
So,There are 67 different ways to complete the race.