A problem of combinatorics

 Posts: 1
 Joined: Sat Dec 12, 2015 8:48 pm
A problem of combinatorics
There are 36 participants at a BDMO event . Some of the participants shook hands with each other . But no two participants shook hands with each other more than once . Each participants recorder the number of handshakes they made . It was found that no two participants with the same number of handshakes made , had shaken hands with each other . Find the maximum number of handshakes at the party with PROOF .(when 2 participants shook hands with each other ,this will be counted as one handshake )
 seemanta001
 Posts: 13
 Joined: Sat Jun 06, 2015 9:31 am
 Location: Chittagong
Re: A problem of combinatorics
A person can shake hands with others for maximum $35$ times.
Suppose $A_1$ shook hands 35 times with $A_2$,$A_3$....$A_{36}$.
Then $A_2$ can shake hands maximum $34$ times.As he has already shaken hands with $A_1$, suppose he hasn't shaken hands with $A_3$.Similarly $A_3$ hasn't shaken hands with $A_2$ and can shake hands maximum $34$ times.
Thus the maximum number of handshakes in that party will be:
$$35+(2 \times 34)+(3 \times 33)+(4 \times 32)+.....+(8 \times 28)=1092$$.
Suppose $A_1$ shook hands 35 times with $A_2$,$A_3$....$A_{36}$.
Then $A_2$ can shake hands maximum $34$ times.As he has already shaken hands with $A_1$, suppose he hasn't shaken hands with $A_3$.Similarly $A_3$ hasn't shaken hands with $A_2$ and can shake hands maximum $34$ times.
Thus the maximum number of handshakes in that party will be:
$$35+(2 \times 34)+(3 \times 33)+(4 \times 32)+.....+(8 \times 28)=1092$$.
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: A problem of combinatorics
The actual solution to this problem is like this
1 can shook hands with a max. of 35 people.
Then 2 can shook hands with a max. of 34 people (this is the total handshake made including the handshake with the 1st person).
Then 3 with max. 33, 4 with 32, 5 with 31, 6 with 30, 7 with 29 and 8 with 30 people.
Now, if everybody shook hands with each other, then total no. of handshake would be= 36C2= 630.
So now let us calculate how many handshake didn't occur
2C2+3C2+4C2+5C2+6C2+7C2+8C2=84
Therefore, max. no. of handshake was= 63084=546.(ans)
1 can shook hands with a max. of 35 people.
Then 2 can shook hands with a max. of 34 people (this is the total handshake made including the handshake with the 1st person).
Then 3 with max. 33, 4 with 32, 5 with 31, 6 with 30, 7 with 29 and 8 with 30 people.
Now, if everybody shook hands with each other, then total no. of handshake would be= 36C2= 630.
So now let us calculate how many handshake didn't occur
2C2+3C2+4C2+5C2+6C2+7C2+8C2=84
Therefore, max. no. of handshake was= 63084=546.(ans)

 Posts: 11
 Joined: Tue Jun 16, 2015 5:11 am
 Location: Barisal, Bangladesh
Re: A problem of combinatorics
I didn't get the last line. please explain.Ragib Farhat Hasan wrote:The actual solution to this problem is like this
1 can shook hands with a max. of 35 people.
Then 2 can shook hands with a max. of 34 people (this is the total handshake made including the handshake with the 1st person).
Then 3 with max. 33, 4 with 32, 5 with 31, 6 with 30, 7 with 29 and 8 with 30 people.
Now, if everybody shook hands with each other, then total no. of handshake would be= 36C2= 630.
So now let us calculate how many handshake didn't occur
2C2+3C2+4C2+5C2+6C2+7C2+8C2=84
Therefore, max. no. of handshake was= 63084=546.(ans)

 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: A problem of combinatorics
Hint
Answer

 Posts: 17
 Joined: Wed Mar 07, 2018 11:35 pm
 Contact:
Re: A problem of combinatorics
36×35/2 =630

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: A problem of combinatorics
I meant to say that subtracting the "no. of handshakes that couldn't occur" (to satisfy the given condition) from the max. no. of handshakes that "could've occurred if there wasn't any condition" provides the final answer, that is 546.Golam Musabbir Joy wrote: ↑Mon Aug 08, 2016 12:45 amI didn't get the last line. please explain.Ragib Farhat Hasan wrote:The actual solution to this problem is like this
1 can shook hands with a max. of 35 people.
Then 2 can shook hands with a max. of 34 people (this is the total handshake made including the handshake with the 1st person).
Then 3 with max. 33, 4 with 32, 5 with 31, 6 with 30, 7 with 29 and 8 with 30 people.
Now, if everybody shook hands with each other, then total no. of handshake would be= 36C2= 630.
So now let us calculate how many handshake didn't occur
2C2+3C2+4C2+5C2+6C2+7C2+8C2=84
Therefore, max. no. of handshake was= 63084=546.(ans)

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: A problem of combinatorics
Read the problem again.

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: A problem of combinatorics
The correct answer of this problem is 546, and it is an established solution. This problem actually appeared in the National Math Olympiad 2015 in secondary and higher secondary categories.
Moreover, read the question again. It says that no one shook hands with anyone more than once. In absence of any other conditions (though the problem states another condition), the max. no of handshake will be 36C2=630.
But under the given conditions, the answer is 546.
I hope you find where you made mistakes and correct them.

 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: A problem of combinatorics
This is BdMO National Secondary & Higher Secondary $2015$ problem no $4$. My solution was wrong