A Handshaking Problem

This is a direct application of combination : there are 10 people, and they shake hands with each other. Each handshake requires 2 people.

Writing it in terms of combination, we have $^{10} C_2 = \dfrac{10!}{2!(10-2)!} =\dfrac{10!}{2!8!} = \dfrac{10\times9\times 8!}{2 \times 8!} = \dfrac{10\times9}2 = \boxed{45}$

For the solution we should realise that in a group of n people first person will shake hands with everyone except himself so there will be n-1 handshakes. For the second person he wont shake hands with himself and the first person again so we have n-2 handshakes. For third n-3 and so on which forms an arithematic progression with n terms the common difference d=-1. Let the first term be a and the total sum as 45. As we know the first person will shake n-1 hands a=n-1. Using the formula for total sum we get 45=n/2 ( 2a+d(n-1))
Substitute a and d to get
45=n/2 (2(n-1)-1(n-1))
45=n/2 (2n-2-n+1)
45=n/2 (n-1)
Solve this and you get n=10

$\sum _{ k=1 }^{ n-1 }{ k } = 45$

$n-1=9$

$\therefore \quad n=10$

OR

${ _{ n }{ C }_{ 2 } }=45$

$\frac { { _{ n }{ P }_{ 2 } } }{ 2! } =\frac { n! }{ (n-2)!\cdot 2! } =45$

${ _{ n }{ P }_{ 2 }=\frac { n! }{ (n-2)! } }=90$

$90(n-2)!\quad =\quad n!$

$n\cdot (n-1)\quad =\quad 90$

${ n }^{ 2 }-n-90=0$

$n=\frac { 1\pm \sqrt { 1-4\cdot 1\cdot -90 } }{ 2 } =\frac { 1\pm \sqrt { 1+360 } }{ 2 } =\frac { 1\pm 19 }{ 2 }$

$n=10\quad or\quad -9$

But since there can't be negative number of people,

$\therefore \quad n=10$

Given : The number of handshakes = 45 We know that, If 'n' is the number of nodes in a graph then the number of paths,x, between two nodes so that each node gets connected to every other node is given by

x= $\frac{ n(n-1) }{ 2 }$

Thus, $\frac{ n(n-1) }{ 2 }$ =45

       => n(n-1) = 90

       => n=10.

Therefore, 10 People were there in the room.

as for every handshake 2 persons are required hence a combination of 2 person is made. it's sol will be nC2=45 on solving we get, n=10

I will try to PROVE the formula to get the answer, so I think it's easier to compreend ^^

We have "x" people handshaking their hands. Observing each person we can say:

Each person will handshake x-1 times. It's easy to think: If you have 3 people, 1 guy will handshake his hand with 2 others If you have 4 people, 1 guy will handshake his hand with 3 others If you have 5 people, 1 guy will handshake his hand with 4 others Than... If you have "x" people, 1 guy will handshake his hand with "x-1" others

But you can't simply sum 'x' times 'x-1' handshakes and say it's the total ( for example, you can't say that 4 people will handshake their hands on a total of 4 + 4 + 4 or 4*3= 12, when the correct is 3 + 2 + 1, that is 6 handshakes). That happens because when you count for one guy, the next one will do x-2 handshakes PLUS the handshake given by the first guy that isn't counted, because it will be a repeated handshake. For the third guy, it will be x-3 handshakes, as we are NOT counting the handshakes given from the first AND the second guy. For the forth we have x-4 and it keeps going, until we reach the "(x-1)" guy, that will do x - (x-1) (that is equal to 1), not counting the others. If we kept kept going counting the last guy ( 'x'), we would have x-x handshakes by him that is equal to 0 (again, not counting the handshakes from the others), so it's irrelevant count him.

So our formula to solve the total of handshakes (H) given by people (x) will be:

H = x - 1 [from first guy] + x - 2 [from second guy] + x - 3 [from third guy] + ... + x - (x-3) [ from (x-3) guy] + x - (x-2) [ from (x-2) guy] + x - (x-1) [ from (x-1) guy]

H = x - 1 + x - 2 + x - 3 + ... + x - (x-3) + x - (x-2) + x - (x-1)

We can do some math tricks here to simplify this formula and discover what we want.

Let's think that there is 7 people. By the formula...

H(7) = x - 1 + x - 2 + x - 3 + x - 4 + x - 5 + x - 6 + x - 7

Notice the numbers of x's we have. It's exactly 7! So we can rewrite it as:

H(7) = 7x - 1 - 2 - 3 - 4 - 5 - 6 - 7 As x = 7 H(7) = 7*7 - 1 - 2 - 3 - 4 - 5 - 6 - 7

Now another observation. If we sum: 1 + 6 2 + 5 3 + 4 We will Have the number 7! Notice we have 3 sums here. So we cam rewrite the equation as:

H(7) = 7 7 - 3 7 - 7 H(7) = 7 7 - 4 7

Evidanciating 7 in ... H(7) = 7(7-4) H(7) = 7*3

But 3 = 6/2 so... H(7) = 7*6/2

That's the formula for 7. For any other number we can rewrite it as:

H(x) = x * (x-1)/2

We found it! As we want to find x for H = 45, we do:

45 = x * (x-1)/2 90 = x^2 - x x^2 - x - 90 = 0

x = [ 1 +- sqrt ( 1 + 4*90) ]/2

x = ( 1+- sqrt361 )/2

x = (1+ 19) /2 (we will not count the negative as we can't have negative people)

x = 20/2 = 10

If there are $n$ people standing in a circle, and everyone shakes everyone else's hand on their turn. That makes a total of $n \times (n-1)$ handshakes. We need to divide this number by 2 because there are 2 handshakes between $A$ and $B$ , when $A$ goes around and when $B$ goes around. Therefore the correct number of handshakes between $n$ people is given by:

$\frac{n \times (n - 1)}{2}$

Setting this equal to 45 we get:

$45 = \frac{n \times (n - 1)}{2}$

$90 = n^2 - n$

$0 = n^2 - n - 90$

Using the quadratic formula we get:

$n = \frac{-1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-90)}}{2 \cdot 1}$

$n = 10$ or $n = -9$

$n = -9$ does not make sense in the context of this problem, so $n = 10$ is the solution to the problem.

Familiarize this with the Barclay Premier League.

The PL consists of 20 teams and they play one another two times in one season.

A PL fan will know that the number of matches in 1 season is 380.

But since a team plays one another 2 times (home-away). So, in the same way, 20 team captains will make 190 hand shakes if they were in the situation in the problems. So, what do these numbers tell us?

No. of matches : 20×19

No. of handshakes: $\frac {20×19}{2}$

So, generally, no of handshakes = $\frac {n (n-1)}{2}$ when $n$ is the number of people.

So, apply this to the question, solve a quadratic equation, eliminate the impossible and we will get 10 the end of the page

Note: This solution is bad compared to others. It requires you to be a football fan.But generally, you can figure out the general form by starting with small integers.

To answer this question you can reverse engineer the following question:

How many lines can I draw between the points of an $n$ -gon. Think of the people shaking hands as vertices and each one connects (shakes hands) exactly one time. (a triangle would be three people, a square four people, and so on)

An easy way to calculate this is to first total the number of sides of your $n$ -gon. Let's use a square as an example. Thus our total number of sides is $4$ . Now add that number to the number of total diagonals you can make connecting vertices through the middle of your square. The formula for this is $n(n-3)/2$ where $n$ is the total number of sides your $n$ -gon has.

So if we have four people the total handshakes that occur is equal to

$n+n(n-3)/2=4+4(4-3)/2=4+2=6$

Now let's extend this equation to our question. We know $45$ total handshakes occurred (think of this as we have an $n$ -gon with $45$ edges connecting all the vertices).

Thus

$\begin{aligned} n+n(n-3)/2&=45\\ 2n+n(n-3)&=90\\ 2n+n^2-3n&=90\\ n^2-n&=90\\ n^2-n-90&=0\\ (n+9)(n-10)&=0 \end{aligned}$

You get two solutions to the quadratic equation $n=10$ and $n=-9$ . Discard your negative solution and you have your answer, in this case $10$ .

The question can be interpreted as describing a complete graph with 45 edges, like this: The answer we want is the number of nodes in this graph. Although we could simply count the nodes in this graph, we could also solve for $n$ by setting this formula (the relationship between the number of edges and nodes) - ${n(n-1)/}{2}$ - equal to 45. Then, we can simply multiply by two to obtain $n(n-1)=90$ . We can see that we need two successive positive integers whose product is 90. These can be quickly found to be 9 and 10, and 9 is simply 10 minus 1, therefore 10 is the answer.

If there are n persons present, each of them will shake hands with the other (n-1) persons, giving a total of n(n-1) handshakes. This method would count each handshake twice (once each from the perspective of the two persons involved.) So n(n-1) = 2x45 = 90 and it is easy to see that n = 10.

I put N persons in the vertices of a polygon, so the total handshake is equal to the number of sides (S = N vertices) + the number of diagonals (D)...

$D = \frac{N(N-3)}{2} \Rightarrow S + D = N + \frac{N(N-3)}{2} = \frac{N(N-1)}{2} = 45 \Rightarrow N = \boxed{10}$

the negative solution is obviously not valid.

The result can be obtained by simple reasoning, without reference to formulae, or graph node theory etc. One person shakes hands with (n-1) people. n persons shake hand with n(n-1) people. But half these handshakes can be ignored as repeats. Hence there are n(n-1)/2 handshakes = 45. Hence n = 10

Try the triangle number formula backwards. Instead of n(n+1)/2

You can multiply the number of handshakes by 2 and find 2 consecutive numbers that multiply to make the number. In this case, we multiply 45 by 2 to make 90 and find 2 consecutive numbers that multiply to make 90. In this case,9x10. We then choose the higher number and that's your answer

no. of persons =n and according to question nc2=45

(x-1)+(x-2)+….+[x-(x-1)]=45 => x=10

The explanation:

[n.(n+1)]/2=45

n=x-1

x²-x-90=0

(x-10).(x+9)=0

X=10

since 45 can be written as 1+2+3+4+5+6+7+8+9.

no one shake hands with them itself.

so no of people=10.

n(n-1)/2 = 45

n2 - n = 90

n2 - n - 90= 0 (n-10) (n+9) = 0

Therefore n =10 or n= -9 (not applicable) Then n= 10

Its easy.... if 45 handshakes are occurring, then there are 90 hands meeting. Assuming every person shakes only one hand, every person will shake (n-1) hands. And since there are n number of people we have (n-1)n=90 which gives n^2-n-90=0 solving this we get, n=10 or n=-9

Obviously n=10

thus n=10

suppose x people were present. so, xC2 = 45; or, x! / (x-2)! * 2! = 45; or, x* (x-1) * (x-2)! / (x-2)! = 45 * 2!; or, x*(x-1) = 90;

solving this equation, we have x=10 and x=-9; since the question mentions that x>0; so x=10 (answer)

nC2 number of shake hands occur which are given to be equal to 45. on solving we get a quadratic...which has feasible solution as 10.

so according to the problem one person shakes hand with all others persons. if there is two person then 1 handshake occurs for each . if 3 person then 2 . and like the these if 10 then 9. now the number of handshake is 1+2+3+4+5+6+7+8+9=45. so there are 10 persons.

nC2 = 45

n!/2!(n - 2)! = 45

n! = 2!(n-2)45

n(n-1)(n-2)!=90(n-2)!

n(n-1)=90

n(n-1)=9×10

Here it is clear that n=10 and n-1=9

9+8+7+6+5+4+3+2+1=45 So there are 10 people in there.

solve for the nc2 and you will get the answer!!!