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### A problem of combinatorics

Posted: Wed Dec 23, 2015 1:25 am
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 )

### Re: A problem of combinatorics

Posted: Thu Jan 14, 2016 12:39 am
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:
$$35+(2 \times 34)+(3 \times 33)+(4 \times 32)+.....+(8 \times 28)=1092$$.

### Re: A problem of combinatorics

Posted: Thu Jan 14, 2016 7:32 pm
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= 630-84=546.(ans)

### Re: A problem of combinatorics

Posted: Mon Aug 08, 2016 12:45 am
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= 630-84=546.(ans)
I didn't get the last line. please explain.

### Re: A problem of combinatorics

Posted: Sun Feb 18, 2018 10:51 am
Hint

### Re: A problem of combinatorics

Posted: Fri Mar 09, 2018 5:49 pm
36×35/2 =630

### Re: A problem of combinatorics

Posted: Mon Sep 03, 2018 1:22 am
Golam Musabbir Joy wrote:
Mon Aug 08, 2016 12:45 am
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= 630-84=546.(ans)
I didn't get the last line. please explain.
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.

### Re: A problem of combinatorics

Posted: Mon Sep 03, 2018 1:31 am
Mathlomaniac wrote:
Fri Mar 09, 2018 5:49 pm
36×35/2 =630

### Re: A problem of combinatorics

Posted: Mon Sep 03, 2018 1:33 am
samiul_samin wrote:
Sun Feb 18, 2018 10:51 am
Hint
This is BdMO National Secondary & Higher Secondary $2015$ problem no $4$. My solution was wrong