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Relationship between ppl and # of handshakes

  1. Feb 7, 2005 #1
    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 dont 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 im 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: Im 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 :cry: STEP by STEP PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
     
    Last edited: Feb 7, 2005
  2. jcsd
  3. Feb 7, 2005 #2
    I really need help I know this question is long but can someone plz reply! :blushing:
     
  4. Feb 7, 2005 #3
    Have you learned about recurrsive relationships yet?
     
  5. Feb 7, 2005 #4

    dextercioby

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    How about combinatorics...?

    Daniel.

    P.S.Is that a question?Judging by the length,i though there were 100... :tongue2:
     
  6. Feb 8, 2005 #5

    cepheid

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    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.
     
  7. Feb 8, 2005 #6

    HallsofIvy

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    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.)
     
    Last edited: Feb 8, 2005
  8. Feb 8, 2005 #7

    cepheid

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    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

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

    But now I'm confused. I thought it was 1 + 2 + 3 + ... + (n-1)
    ?
     
  9. Feb 8, 2005 #8

    HallsofIvy

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    Typo- I'll edit it.
     
  10. Feb 8, 2005 #9
    lol i can barely say that word let alone know what it means.
     
  11. Feb 8, 2005 #10
    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:
     
  12. Feb 9, 2005 #11

    HallsofIvy

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    You've been given the formula that answers the last question. And that formula is NOT linear.
     
  13. Feb 9, 2005 #12
    Aisha, the equation you have been given is [tex]\frac{n(n - 1)}{2}[/tex]

    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 ©)
     
    Last edited: Feb 9, 2005
  14. Feb 10, 2005 #13
    [tex] \frac {n^2 -n } {2} [/tex]? why do i want another equation? Am i trying to get a linear relationship? is this linear now?
     
  15. Feb 10, 2005 #14

    learningphysics

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    What type of equation is the above? You've worked with these kinds of equations... they're not linear but....
     
  16. Feb 10, 2005 #15
    Oh isn't this a quadratic relationship? Can you also check my other questions please?
     
  17. Feb 10, 2005 #16
    The penny has dropped. :biggrin:
    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 ©)
     
  18. Feb 11, 2005 #17
    I actually believe I did this very same problem recently, only with slightly more people. Anyway, this is my guess:
    [tex]C_{5}^{10} = \frac{10!}{5!(10-5)!} = 252[/tex]
    Correct me if I'm wrong, though.
     
  19. Feb 11, 2005 #18
    Well I thought the same but [tex]^{10}C_{0}[/tex] is equal to [tex]^{10}C_{10}[/tex] so the number of handshakes would not increase.

    The Bob (2004 ©)
     
  20. Feb 11, 2005 #19

    learningphysics

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    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.
     
    Last edited: Feb 11, 2005
  21. Feb 25, 2005 #20
    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 plz help me? :redface:
     
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