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Combinatorics help....

  1. Oct 5, 2015 #1
    1. The problem statement, all variables and given/known data
    I hope someone can help me as it is urgent...
    We randomly assign 40 people (20 girls and 20 boys) into 20 double rooms (we can assume that all room assignments are equally likely). What’s the probability that there will be no mixed sex rooms?

    2. Relevant equations


    3. The attempt at a solution
    I wrote:
    P = {20!/[(2!)*10] } / {40!/[(2!)*10] }
    = (20!) / (40!)
    = 2.98*10^-30

    I have no confidence to my answer since I am still very unfamiliar with combinatorics.I think it is the probability of choosing 2 girls out of 20 girls each time,then 2 from 18 .....and so on.Dividing this with the denominator,which is choosing 2 people out of 40.....2 out of 38....until 2 out of 20 and so you get 2!*10

    Please help...Thanks so much
     
  2. jcsd
  3. Oct 5, 2015 #2

    Ray Vickson

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    Let's do a simpler case: 4 rooms and 8 people (4 boys, 4 girls).
    The probability of 2 girls in room 1 is
    [tex] p_1 = \frac{4}{8} \cdot \frac{3}{7} = \frac{3}{14} [/tex]
    After those 2 girls have been chosen, the probability of 2 girls in room 2 is
    [tex] p_2 = \frac{2}{6} \cdot \frac{1}{5} = \frac{1}{15} [/tex]
    So, the probability of 2 girls in rooms 1 and 2 is ##P = p_1 p_2 = 1/225##.
    However, this 2-girls in 2-rooms probability is the same for any pair of rooms, so the answer is the above, times the number of pairs of 2 rooms from 4. That is,
    [tex] \text{Answer} = {}_4C_2 \cdot P = \frac{6}{225} \doteq 0.0266666666 [/tex]
     
  4. Oct 5, 2015 #3
    Sorry,I am stuck on the first step...
    p1=4/8⋅3/7=3/14
    Why do we have to multiply in this way?
     
  5. Oct 5, 2015 #4

    haruspex

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    We are choosing two people for room 1. First choice is from 4 and 4, so odds of choosing a girl is 4/8. Having chosen a girl, odds of choosing a second girl from the remaining 3 girls and 4 boys is 3/7. So probability of choosing two girls for room 1 is (4/8)(3/7).
     
  6. Oct 5, 2015 #5
    Following your logic,the probability of two girls in 1st room = (20/40)(19/39) = 0.243589....
    the probabiility of two girls in 2nd room = (18/38)(17/37) = 0.21763....
    As the no of girls available is decreasing,and as we are distributing the students over a number(20) of rooms,so I guess the probability of each room should not be equal.I mean....the probability should decrease as the distribution goes on?Just like
    (20/40)(19/39)(18/38)(17/37).................(1/21)
     
  7. Oct 5, 2015 #6

    Ray Vickson

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    If we label the evens as E1 = {2 girls in room 1} and E2 = {2 girls in room 2}, we want P(E1 & E2). If you have studied conditional probability, you will realize that P(E1 & E2) = P(E1) * P(E2 | E1), where the second factor is the conditional probability of E2, given that E1 has occurred. The 4/8 and 3/7 are just P(E1) and P(E2 | E1).

    Another way to see this is to look at a "sample space" for the whole problem. We have 8 people, labeled 1-8. A random choice of people to rooms consists of taking a random permutation of the numbers fro 1 to 8, then taking the first two and putting them in room 1, the next two in room 2, etc. Altogether, there are 8! permutations. In how many permutations do we have 2 girls in each of rooms 1 and 2? Say the girls are numbered 1-4. The number of permutations of 1-8 where 1-4 are in rooms 1-2 is 4*3*2*1*4!. Here, 4 is the number of ways of choosing the first girl in the list, then 3 is the number of ways of choosing the second girl, etc. After choosing 4 girls, there are 4! ways of permution the 4 boys remaining. So, the probability of 4 girls in the first two rooms is P = N/D, where N = number of 'favorable' permutations = 4*3*2*1*4!, and N = total number of permutations = 8! = 8*7*6*5*4!. Thus, P = N/D = (4*3*2*1)/(8*7*6*5) = (4/8)(2/7)(2/6)(1/5), just as given before.
     
  8. Oct 5, 2015 #7

    haruspex

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    That is correct. However, this only gives the odds of one particular way in which there are no mixed sex rooms. What do you need to multiply by to get the final answer?
     
  9. Oct 5, 2015 #8
    I think (20/40)(19/39)(18/38)(17/37).................(1/21) represent the probability of choosing 20 girls from 40 people,so it should be the answer?Because boys are not included,so there is no mix gender room?
     
  10. Oct 6, 2015 #9

    haruspex

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    Which rooms are the girls in?
     
  11. Oct 6, 2015 #10
    Um....I just have a guess
    (the possible combinations of 10 rooms out of 20rooms)(probability of choosing all girls out of 40).The first one represents 20 people are chosen when you pick 10 rooms out of 20 rooms,the latter represents the probability to pick 20 girls,1 by 1,out of 40 people?I am a bit fussed.
     
  12. Oct 6, 2015 #11

    haruspex

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    I think you have it. The point is that the way you counted the possibilities for having girls only in 10 rooms was as though it was a specific 10 rooms, but it could be any 10 rooms.
     
  13. Oct 6, 2015 #12
    Ok, thanks a lot. To me,probability is really confusing since my brain cant figure out what's going on: ( .......Whenever I am going solve, my mind just like getting entangled.
     
  14. Oct 6, 2015 #13

    vela

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    Another approach which may have been along the lines of what you were initially thinking, applied to Roy's example:

    There are ##{}_4C_2## ways to fill room 1 with two girls, and ##{}_2C_2## ways to fill room 2 with the remaining two girls. Similarly, there are ##{}_4C_2## ways to fill room 3 with two boys, and ##{}_2C_2## ways to fill room 4 with the remaining two boys. If you work this out, you get
    $$\left( \frac{4\times 3}{2 \times 1} \cdot \frac{2 \times 1}{2 \times 1} \right)^2 = \left(\frac{4!}{2!^2}\right)^2 = 36.$$ Since there are ##{}_4C_2 = 6## ways to assign the rooms to the girls, there's a total of ##36 \times 6 = 216## ways to put pairs of girls into the rooms.

    To fill the rooms with no conditions, you can choose 2 of 8 to put into the first room, 2 of the remaining 6 in the second room, and so on, to get
    $${}_8C_2 \cdot {}_6C_2 \cdot {}_4C_2 \cdot {}_2C_2 = \frac{8\times 7}{2 \times 1} \cdot \frac{6 \times 5}{2 \times 1} \cdot \frac{4\times 3}{2 \times 1} \cdot \frac{2 \times 1}{2 \times 1} = \frac{8!}{2!^4} = 2520.$$ The probability is then 216/2520 = 3/35.

    In your initial attempt, it looks like some errors arose because you didn't count how to arrange the boys.

    Note Ray made a mistake multiplying ##p_1## and ##p_2## in post 2 and should've gotten ##1/70##, which when multiplied by 6 gives you ##3/35##.
     
    Last edited: Oct 6, 2015
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