Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By richard1234

Math Help - Probability of C+A+R+D=35

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    South Korea
    Posts
    17

    Probability of C+A+R+D=35

    I was asked to solve the below. Could I get help?

    Four letters of C, A, R and D have its own natural number: 1,2,3,4,5,.... Sum of the four letters, C+A+R+D=35.

    A. Total number of possible pairs?
    B. Probability of C=32 or A=32 or R=32 or D=32 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Probability of C+A+R+D=35

    A: Do you know the "balls and boxes" or "divider" technique?

    Basically, we subtract 1 from each number so that C,A,R,D are nonnegative integers with C+A+R+D = 31. Now we have 31 units and three "dividers" to split with. Any permutation of the 31 units and three dividers results in four numbers C,A,R,D such that C+A+R+D = 31. The total number of ways is \binom{34}{3} = \frac{34!}{31!3!}.

    B: Let C = 32, find the number of ways such that A+R+D = 2. Multiply by 4 to account for A = 32, etc.
    Thanks from tykim
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    South Korea
    Posts
    17

    Re: Probability of C+A+R+D=35

    Thanks richard1234 for the quick reply.
    Very helpful to me.

    TYKim
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2012
    From
    South Korea
    Posts
    17

    Re: Probability of C+A+R+D=35

    Dear Richard1234,

    I can not get it with problem B. Could you give me further information?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Probability of C+A+R+D=35

    Okay. Suppose C = 32. How many possible ways are there such that A+R+D = 3 (sorry, not 2)? This one's pretty easy to brute force.

    Then suppose A = 32. You want the number of ways such that C+R+D = 3. However this is the same as with the C = 32 case. That's why you multiply by 4.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 11th 2012, 05:42 AM
  2. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  3. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum