# if there are 24 students. how many handshakes would there be? grade 7?

### 4 Answers

- PuzzlingLv 75 years agoFavorite Answer
If you imagine each of the 24 people, they must shake hands with 23 other people.

For example, let's imagine you have people named A, B, C, ..., T, U, V, X.

A would have to shake hands with B, C, D, ..., T, U, V, X.

B would have to shake hands with A, C, D, ..., T, U, V, X.

C would have to shake hands with A, B, D, ..., T, U, V, X.

etc.

The only problem is you would double-count every single handshake. You'd not only count A shaking hands with B, but B shaking hands with A.

So the net result is you need to divide that by 2.

In other words:

(24 x 23) / 2

= 12 x 23

= 276 handshakes

Summary:

With n people, you need n(n-1)/2 handshakes. Specifically with 24 people, there would be (24 x 23)/2 = 276 handshakes

- MichaelLv 75 years ago
Well,

first shakes 23 hands

second shakes 22 hands

etc.

answer = 1 + 2 ... + 23

= 23*24/2

= 276

276 shakes !

PS: wanna know how to get the result ?

in the 19th century, a teacher asked his children, so to have a quiet moment, to add all numbers from 1 to 100

he was surprised as, after a short time, the answer came : "5050, Sir !"

the name of the child was Karl-Friedrich GAUS !

Sn = 1 + ... + (n-1) + n

Sn = n + ....+ 2 ... + 1

add vertically :

2Sn = (n+1) + (n+1) + (n+1) + ... + (n+1) <--- n terms

therefore :

Sn = n(n+1)/2

for 100 : S100 = 100*101/2 = 5050

et voilà !! ;-)

hope it' ll help !!

- ThomasLv 75 years ago
The first person must shake hands with 23 others, and one of those must now shake hands with 22 others, and the next person must shake hands with 21 others...etc, etc This can be done using arithmetic sequences...

The sequence starts with 23 decreases by 1 each time and the last term is 1 so

an=23-1(n-1) or

an=23-n+1

an=24-n and our last term, since we need this for our number of terms..

1=24-n

n=25

The sum of handshakes is the arithmetic sum...

s(n)=(2an+dn^2-dn)/2

s(25)=(2*23*25-25^2+25)/2

s(25)=275