Relationship between ppl and # of handshakes

[aisha]

Major Handshake Help Please!

QUESTION 1 :

said Determine the number of handshakes required in a room that contains 5 people, if each person shakes hands with every other person once? (Hint: Introduce people to the room one at a time in order to develop the concept of an arithmetic series)

MY ANSWER 1:

I got 10 handshakes with 5 people, but I don't know how to write this as a series, I used a diagram.

QUESTION 2:

The next question said there are 6 people how many handshakes will occur now? And predict number of handshakes with 10 people, a chart was given to organize the work.

#people -- #handshakes -- Total
0 ------------0------------0
1-----------0+1-----------1
2-----------0+1+2---------3
3
4
5

MY ANSWER 2: For 6 people I got 15 handshakes for 10 people I predicted 45 hanshakes. There is a problem with the chart given above which I noticed, 1 person cannot shake hands alone so the total for 1 person should be 0.

QUESTION 3: Identify the type of relationship between the number of people in the room and the total number of handshakes.

MY ANSWER 3: I wrote Linear but I am not sure.

QUESTION 4: Find an equation that best models this relationship. It says the #handshakes if there are 5 ppl in room can be found by adding 0+1+2+3+4, the numbers 0,1,2,3,4 are terms of an arithmetic sequence. Explain why.

MY ANSWER 4: I am not sure but I think the numbers are terms of an arithmetic sequence because there is a common difference between two successive terms.

QUESTION 5: It tell you if there are 10 people then you can do this

S10 = 0+1+2+3+4+5+6+7+8+9
S10=9+8+7+6+5+4+3+2+1+0

Adding these two rows we get:

2S10 = 9+9+9+9+9+9+9+9+9+9
2S10=10(9)
therefore S10=90/2=45

Using a similar method find the total number of handshakes required if there were 20 people, or 50 people in the room. Find the total number of handshakes required if there are n people in the room.

MY ANSWER: I'm not sure on how to solve this question, can't I use the equation I was to form in question 4?

PLEASE HELP ME WITH THIS VERY LENGTHY PROBLEM STEP by STEP PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

 

[aisha]

I really need help I know this question is long but can someone please reply!

 

[MathStudent]

Have you learned about recurrsive relationships yet?

 

[dextercioby]

How about combinatorics...?

Daniel.

P.S.Is that a question?Judging by the length,i though there were 100... :tongue2:

 

[cepheid]

Question 1: I think that the point of introducing the people to the room one at a time is so that each person in the room only has to shake hands with the new guy, every time you introduce a new guy to the room. So, with an empty room, you introduce one guy, and he has no one to shake hands with:

no. of handshakes: 0

Introduce a second guy, and he shakes hands with the first:

no. of handshakes: 1

Introduce a third guy, and he shakes hands with each of the first two:

no. of handshakes: 2

Introduce a fourth guy, and he shakes hands with each of the first three:

no. of handshakes: 3

Introduce a fifth guy, and he shakes hands with each of the first four:

no. of handshakes: 4

Total number of handshakes: 0 + 1 + 2 + 3 + 4 = 10

So with this exercise, I've shown you how to develop the arithmetic series of #1. I've given you what you should need to correct the table in #2. I've confirmed the half of the answer that you gave to #4 (what about the other half? What is the general formula? If I had six people, I'd add up the numbers from 0 (or 1) to 5, If I had 7 people, I'd add up 1 to 6, If I had 20 people, I'd add up 1 to 19. If I had "n" people, I'd add up...?

^^^If you do see what the general formula is, then to answer you question in #5, yes, you could use it to answer #5, but what if you had 100 people? Or 1000 people? Wouldn't it be pretty tedious to sit there and add up the numbers 1 + 2 + ... + 998 + 999? Number five is showing you an even better method that works for all arithmetic series (if you adjust for common differences other than 1). It's a technique that was developed by Gauss. It's really neat. Take the time to make sure you understand it.

 

[HallsofIvy]

That looks like the hard way to me!

Everyone in the room has to shake hands with everyone else: there are n people and each must shake hands with the other n-1 people: n(n-1).

But one handshake counts for two people: n(n-1)/2.

That is, of course, the same as 1+ 2+ 3+...+ (n-1 ).

(Thanks to Cepheid for catching the typo I had in the last line.)

 

[cepheid]

That looks like the hard way to me!

WOW I admire the succintness of your solution... :rofl: Although I was also trying to develop it along the lines of the hint in #1

Everyone in the room has to shake hands with everyone else: there are n people and each must shake hands with the other n-1 people: n(n-1). But one handshake counts for two people: n(n-1)/2.

Okay. Yeah. Wait, isn't that the general formula she was intended to derive in #5?

That is, of course, the same as 1+ 2+ 3+...+ (n-2 ).

But now I'm confused. I thought it was 1 + 2 + 3 + ... + (n-1)
?

 

[HallsofIvy]

Typo- I'll edit it.

 

[aisha]

How about combinatorics...?

Daniel.

P.S.Is that a question?Judging by the length,i though there were 100... :tongue2:

lol i can barely say that word let alone know what it means.

 

[aisha]

Ok I want to know was my answer for number 3 correct?
"Linear relationship between total number of handshakes and number of people."

Also did my answer for number 4 make sense?
"Im not sure but I think the numbers are terms of an arithmetic sequence because there is a common difference between two successive terms."

I understand what Gauss did but for number 5 wouldn't it be easier to just use the equation. n(n-1)/2?
Using this equation for 20 people I got 190 handshakes and for 50 people I got 1225 handshakes. After the example in number 5 showing Gauss' method the question did say use a similar method for 20 and 50 people but y?

IF I have to use GAUSS's METHOD HERE IT IS...

S20=0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19
S20=19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1+0

Adding these two rows we get
2S20=19+19+19+19+19+19+19+19+19+19+19+19+19+19+19+19+19+19+19+ 19

2S20=20(19)

Therefore S20=380/2
=190

OH also the last question said find the total number of handshakes required if there are n people in the room, what do u do? How do u find out? :yuck:

 

[HallsofIvy]

You've been given the formula that answers the last question. And that formula is NOT linear.

 

[The Bob]

Aisha, the equation you have been given is $$\frac{n(n - 1)}{2}$$

This, as HallsofIvy said, is not linear. Try doing something to the equation that will NOT change the end result, e.g. the answer when you put n in, but will make the equation different. Then tell us what it is. It should be more obvious, if you get the hint I am trying to say without giving it away ( ----> '()' ):tongue2:

The Bob (2004 ©)

 

[aisha]

$$\frac {n^2 -n } {2}$$? why do i want another equation? Am i trying to get a linear relationship? is this linear now?

 

[learningphysics]

$$\frac {n^2 -n } {2}$$? why do i want another equation? Am i trying to get a linear relationship? is this linear now?

What type of equation is the above? You've worked with these kinds of equations... they're not linear but...

 

[aisha]

What type of equation is the above? You've worked with these kinds of equations... they're not linear but...

Oh isn't this a quadratic relationship? Can you also check my other questions please?

 

[The Bob]

Oh isn't this a quadratic relationship?

The penny has dropped.

Can you also check my other questions please?

I think it would be better if you repost, in this trhead, all of your answers to the question so it is easy to check for you.

The Bob (2004 ©)

 

[gschjetne]

I actually believe I did this very same problem recently, only with slightly more people. Anyway, this is my guess:
$$C_{5}^{10} = \frac{10!}{5!(10-5)!} = 252$$
Correct me if I'm wrong, though.

 

[The Bob]

I actually believe I did this very same problem recently, only with slightly more people. Anyway, this is my guess:
$$^{10}C_{5} = \frac{10!}{5!(10-5)!} = 252$$
Correct me if I'm wrong, though.

Well I thought the same but $$^{10}C_{0}$$ is equal to $$^{10}C_{10}$$ so the number of handshakes would not increase.

The Bob (2004 ©)

 

[learningphysics]

Oh isn't this a quadratic relationship? Can you also check my other questions please?

Yes it is quadratic.

Your answers look good to me. I'm kind of unsure of why they ask for the equation in #4, and then in #5 ask you to use the summing method... and ask for the number of shakes for n people (since that's already answered using the equation in #4).

Maybe they want: no. handshakes=0+1+2+3+...(n-1) for #4 and n(n-1)/2 for
#5. I'm not sure. Maybe just in case for #4 you should write both:
no. handshakes=0+1+2+3+...(n-1) = n(n-1)/2

You can check the numerical answers yourself using the formula n(n-1)/2. I believe they're all correct.

Be sure to point out the mistake they made in the table for #2.

 

[aisha]

Find the total number of handshakes required if there are n people in the room, I am still not sure how to do this.. can someone please help me?

 

[HallsofIvy]

You have already been told the answer to that in 4 separate posts. Each of the n people must shake hands with n-1 other people: that would be n(n-1). But, since each handshake involves two people, there are really n(n-1)/2.

 

[aisha]

You have already been told the answer to that in 4 separate posts. Each of the n people must shake hands with n-1 other people: that would be n(n-1). But, since each handshake involves two people, there are really n(n-1)/2.

I understand that is the equation but that was another question this one is asking something else