## BDMO NATIONAL JUNIOR 2015/5

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### BDMO NATIONAL JUNIOR 2015/5

In a party, boys shake hands with girls but each girl shake hands with everyone else .If there are 40 handshakes , find out the number of boys and girls in the party ?.........please explain the solution.

aritra barua
Posts: 57
Joined: Sun Dec 11, 2016 2:01 pm

### Re: BDMO NATIONAL JUNIOR 2015/5

Suppose,there are m girls and n boys.Then,1 boy makes m handshakes.For this reason,n boys make in total mn/2 handshakes.Again,1 girl makes m+n-1 handshakes,so m girls make(m^2+mn-m)/2 handshakes.Now,establish the condition,mn/2+(m^2+mn-m)/2=40.....then you may derive m(m+2n-1)=80,then a little checking over the factors of 80 will give your required answer.

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: BDMO NATIONAL JUNIOR 2015/5

The girls have already handshaked with boys.Will they handshake again?Moreover,$1$ boy makes $m$ handshakes.Isn't $n$ boys make in total $mn$ handshakes?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

aritra barua
Posts: 57
Joined: Sun Dec 11, 2016 2:01 pm

### Re: BDMO NATIONAL JUNIOR 2015/5

Hmmm....you are right...I forgot to consider those cases

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: BDMO NATIONAL JUNIOR 2015/5

Let there are $m$ boys and $n$ girls.Thus the number of Boys' handshake = $mn$.And the number of girls' handshake = $\frac{(n-1)^2+(n-1)}{2} = \frac{n^2 - n}{2}$
Thus,$mn + \frac{n^2-n}{2} = 40$
or,$2mn + n^2 - n = 80$
The only solutions of $(m,n)$ is $(40 , 1)$ and $(6,5)$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid