Handshakes

Let us take a series. First man shakes hand with 4 men.Second with 3.Third with 2. Fourth with 1. So 4+3+2+1 =10 (No of hand shakes reduces by every 1 man bcoz they have already shaken hand with the men before them)

5C2 = 10

O problema trata de um combinação de C5,2 = 5!/2!.(5-2)! = 10

Represente as 5 pessoas como A, B, C, D e E. Então, faça todas as correspondências, lembrando que, por exemplo, AB = BA: AB, AC, AD, AE; BC, BD, BE; CD, CE; DE. Ao todo, 4 + 3 + 2 + 1 = 10 apertos de mão.

Let's pretend that first,second, ... person is A,B,C,D,E. Then A shakes hands with B,C,D,E. B shakes hands with C,D,E. C shakes hands with D,E. D shakes hands with E. 10 shakes

4+3+2+1=10

There are five people and two people handshaking at a time. Here, we can use combination. 5C2=5!/(3!*2!). Therefore, the answer is 10.

brute force!

(5*4)/2=10

let 5 people become a, b, c , d, and e. Person a can shake wtih 4 people(b,c,d, e).person b can be shake with 3 people(c,d and e)because b has shook with a before.Person c can only shake with 2 people(d and e) because c has shook with a and b before.Person d can only shake with 1 person(e)becuase d has shook with a , b , and c before.Last,e has shook with everyone and cannot shake himself so the answer is 4+3+2+1=10

its about combination 5C2 = 5!/(5-2)!2! =10 ^^

5 choose 2 = 10

the numbers of handshakes = C(5,2) = 10

Uma pessoa dá 4 abraços e como são 5 pessoas no total, temos 4 vezes 5, porém estamos contando duas vezes os abraços logo devemos dividir por dois

the 1st person will shake 4 people. the 2nd person will shake 3 people. the 3rd person will shake 2. then the 4the person will shake the last person. the total number of shakes are 4+3+2+1=10

O evento (apertar a mão) necessita de duas pessoas, e a ordem em que essas pessoas forem escolhidas não interfere. Logo, se trata de uma combinação de 5 pessoas, duas a duas: 5!/2!.(5-2)!=10

Since there are 5 people, so 1 of the people can only shake hands with 4 people. And since a handshake involves 2 people, so the answer is 5*4/2 = 10

Let the number of friends be n. All n friends shook hands with everybody else and since they did not shake their own hand each individual friend shook hands with $n-1$ other people. So the sum of all such handshakes would be $n(n-1)$ . However, this number is twice too large, because if friend nr 1 shakes hands with friend nr 2 and vice verse than there have been two handshakes between the same two people, so we need to half the result. So the number of handshakes is $(n(n-1))/2$ . Thus the result here is $(5(5-1))/2=20/2=10$ .

combinations of 5 persons. Every person will hand shake with remaining person. So 4+3+2+1=10

1st person shakes for four people.2nd for three people.3rd for 2.4th for 1. i.e. 4+3+2+1 =10 hand shakes

Use the formula of combination(C).

5C2= $\frac{5!}{2!3!}$ =10

Ans=10

5C2 = 10

there's 5 people and each people have 2 hand so 5 x 2 = 10

It is widely known that handshakes has the formula of

((number of people)*(number of people - 1)) / 2

Take 5 as the number of people,

5*(5-1)/2 =25-5/2 =20/2 =10

by cominations nCr

Person = 5 handshakes(n)= ? (5x4)/2= 10

4+3+2+1=10

Given 5 people, start with one person shaking everyone else's hands. Let N be the number of handshakes necessary. To start, N = 4.

Each subsequent person will have one less person to shake hands with, since they've already exchanged farewells. Therefore, each individual's handshake exchange will take the form N-1.

This can be expressed using the summation from i, [1,5], of 5-i.

Imagine the 5 friends standing in a line. The first friend steps out of the line and walks down it shaking the other four friends' hands and then he leaves. The second friend does the same thing but only shakes the remaining three friends' hands. The third shakes the remaining two hands and leaves. The fourth shakes the last hand and leaves. The last friend leaves without shaking a hand because there are none left. The friends shook 4,3,2,1, and 0 hands. 4+3+2+1+0=10

The first person has to shake 5-1=4 people's hands. The second person has to shake 4 people's hands not including the first person so that means that he shakes three new handshakes. The next person has to shake 4 people's hands not including the first two, so that means he shakes 2 new handshakes. The fourth person has to shake 4 people's hands not including the first three, so that means that he shakes 1 new handshake. The last person has already had his hand shaken by everybody else, so he shakes 0 new handshakes. So overall, 4+3+2+1=10 unique handshakes were shaken.

A primeira pessoa vai apertar a mão das outras quatro, a segunda pessoa vai apertar a mão das outras três, pois ja apertou a mão da primeira, a terceira por ter apertado a mão da segunda e da primeira só vai apertar a mão da quarta e da quinta, e assim por diante ate a quinta pessoa. Fazendo assim: (5-1)+(5-2)+(5-3)+(5-4)=10

[n(n-1)]/2

it's just the same as the formula to find the number of segments

the first person shakes his hand with four people the second person shakes his hand with 3 people likewise third person shakes with 2 people and third person with 1 person so to totally there is 4+3+2+1=10 hand shakes

take 5 points and make every possible lines using that five points and then count the number of lines each lines shows one handshake therefore total handshakes are 10

1 = 0, 2 = 1, 3 = 3, 4 = 6, so 5 = 10

It is a basic Combination Problem.As we have 5 people each of them handshakes with every other.We have the formula nCr=n!/(r!*(n-r)!)(Here n=5 and r=2 as two of them shake their hands so we have to find combinations of 2).Applying this we will get the answer as 10

5c2

If the 5 people are $A,B,C,D,E$ then the handshakes are $AB,AC,AD,AE,BC,BD,BE,CD,CE,DE$

"N" people can shake hands with "N-1" people. However, the quantity of shakes have to be divide by 2, because a shake is mutual. So: 5.4=20 --> 20:2 =10

4+3+2+1=10

5C2=10

Let us name the 5 people A, B, C, D, and E. A shakes hands with 4 people -- B, C, D, and E. Person B shakes hands with 3 people -- C, D, and E (he already shook hands with A). C shakes hands with 2 people (D and E) and D shakes hands with 1 person (E). This comes to a total number of 10 handshakes. This method is effective because it avoids over counting.

Note that the number of possible network combinations is: $N=\frac{n(n-1)}{2}$ Which is equivalent to the problem at hand. Note that when $n=5$ , this gives the desired result of $N=10$ .

Alternatively, the sum of numbers below a given case is: $\frac{n(n+1)}{2}$ Note that each person will shake hands with $n-1$ people with $n$ people remaining. Plugging in $n=l-1$ also gives the desired result.

Let the people be vertices of the graph. Since it is said that each shook each others hand. The vertices must be a complete simple undirect graph. Given that there are $5$ vertices, the number of edges that would connect them all is

$N(e) = \dbinom{n}{2}$ so,

$N(e) = \dbinom{5}{2}$

$N(e) = \frac{5!}{(5 - 2)! 2!}$

$= \frac{120}{3!2!} = 10$

Known : total people at Summer Camp = 5 Suppose :Person = n

total of handshake =\binom{n}{k} =\binom{5}{2} = 10