BdMO National Higher Secondary :Problem Collection(2016)

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samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
BdMO National Higher Secondary :Problem Collection(2016)

Unread post by samiul_samin » Fri Feb 16, 2018 12:10 am

You will get question $1,2,3,4,5,6,7,8$ herehttp://matholympiad.org.bd/forum/viewto ... 878#p17476

Now the number 9 and The Final Question

9.
The integral $Z(0)=\int^{\infty}_{-\infty} dx e^{-x^2}= \sqrt{\pi}$

(a)(3 POINTS:)Show that the integral $Z(j)=\int^{\infty}_{-\infty} dx e^{-x^{2}+jx}$
Where $j$ is not a function of $x$,is $Z(j)=e^{j^{2}/4a} Z(0)$

(b)(10 POINTS):Show that,
$\dfrac 1 {Z(0)}=\int x^{2n} e^{-x^2}= \dfrac {(2n-1)!!}{2^n}$
Where $(2n-1)!!$ is defined as $(2n-1)(2n-3)\times...\times3\times 1$

(c)(7 POINTS):What is the number of ways to form $n$ pairs from $2n$ distinct objects?Interept the previous part of the problem in term of this answer.


[It was a 200 number exam,and this is one of the toughest problems.]

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

Re: BdMO National Higher Secondary :Problem Collection(2016)

Unread post by samiul_samin » Fri Feb 16, 2018 12:16 am

My askings.
1.Why so much hard calculas question in this National Olympiad?
2.Is it possible to solve this problem using only the theorems mentioned in our text book of 11-12?
3.From where, in the last equation of (a) a appears(from sky?! :P :?: :?:)
4.How can I interept in (c) .What kind of crossover tactics it is?

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

Re: BdMO National Higher Secondary :Problem Collection(2016)

Unread post by samiul_samin » Fri Feb 16, 2018 12:39 am

Partial Solition of (c)
This solution is taken from Principal and Techniques of Combinatorics
Pick an arbitary element say $x$ from $2n$ numbers.The number of ways to select $x's$ partner say $y$ is $2n-1$ and {$x,y$} forms a pair.
Pick an arbitary element say $z$ from $2n-2$ numbers.The number of ways to select $z's$ partner say $w$ is $2n-3$ and they also form another pair
Counting in this way and applying Multiplication Principle we get the number of ways to make $n$ pairs from $2n$ object is $(2n-3)×(2n-1)...3×1$

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