Bdmo regional 2018 set 1 Higher Secondary 8
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
-
samiul_samin
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
There are $n$ married couples in a party. Each person shakes hands with every person other than his or her spouse. The numbers of handshakes were $3120$. How many couples were there?
Re: Bdmo regional 2018 set 1 Higher Secondary 8
There were $n$ couples that means total $2n$ persons. If each person would shake hands with every other person, the total number of hand shakes would be $^{2n}C_2$. But the total number of hand shakes is $^{2n}C_2 - n$. Because $n$ couple didn't shake hand with their spouse.
Now,
$^{2n}C_2 - n = 3120$
$=> \frac{(2n)!}{2!(2n-2)!} = 3120 + n$
$=> \frac{2n(2n-1)}{2} = 3120 + n$
$=> 2n(2n-1) = 6240 + 2n$
If we solve this, we find that $2n = -78, 80$. But $-78$ is not acceptable.
So, $2n = 80$ $=> n = 40$.
There were $40$ couples in the party...
Now,
$^{2n}C_2 - n = 3120$
$=> \frac{(2n)!}{2!(2n-2)!} = 3120 + n$
$=> \frac{2n(2n-1)}{2} = 3120 + n$
$=> 2n(2n-1) = 6240 + 2n$
If we solve this, we find that $2n = -78, 80$. But $-78$ is not acceptable.
So, $2n = 80$ $=> n = 40$.
There were $40$ couples in the party...