Results 1 to 11 of 11

Math Help - problem on inclusion exclusion

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    12

    Smile problem on inclusion exclusion

    Please describe the way to find the answer for the following question.

    This problem is based on inclusion exclusion principle

    50 cars were assembled in a factory. The options available were radio, AC,
    power steering. It is known that 20 of the cars have radios, 12 of them have
    AC and 10 of them have power steerings. 5 of them have all three options.
    Determine at least how many cars do not have any options at all.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by sudeepmansh View Post
    Please describe the way to find the answer for the following question.

    This problem is based on inclusion exclusion principle

    50 cars were assembled in a factory. The options available were radio, AC,
    power steering. It is known that 20 of the cars have radios, 12 of them have
    AC and 10 of them have power steerings. 5 of them have all three options.
    Determine at least how many cars do not have any options at all.
    A - radio
    B - AC
    C - power steering

    Let n(cars that do not have any options at all)=x

    n(A U B U C) + x = 50
    so, x = 50 - n(A U B U C)
    min(x) = 50 - max(n(A U B U C))

    So what is max(n(A U B U C))?
    n(A U B U C) = n(A) + n(B) + n(c) - n(AB) - n(BC) - n(CA) + n(ABC)

    So all we need is min of each of n(AB), n(BC), n(CA) - So what is it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member oldguynewstudent's Avatar
    Joined
    Oct 2009
    From
    St. Louis Area
    Posts
    244
    Quote Originally Posted by sudeepmansh View Post
    Please describe the way to find the answer for the following question.

    This problem is based on inclusion exclusion principle

    50 cars were assembled in a factory. The options available were radio, AC,
    power steering. It is known that 20 of the cars have radios, 12 of them have
    AC and 10 of them have power steerings. 5 of them have all three options.
    Determine at least how many cars do not have any options at all.
    It doesn't seem that complicated.

    5 cars have all three options.
    If all other cars have only one option apiece until no options are left to install, this gives 20-5=15 with radios only, 12-5=7 with AC only, and 10-5=5 with power steering only.

    Add 5 with all three options, 15 with radio only, 7 with AC only, and 5 with power steering only to get 32 maximum cars with options. 50-32=18, so at least 18 will have no options. There could be more with no options if some of the options are doubled up, but that is beyond the scope of this question.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    12

    Smile

    Thanks a lot for your answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    12

    Thumbs up

    Thanks for your answer, but I am not satisfied with the method. Can anyone please help me to solve using the formula.

    |AUBUC| = |A|+|B|+|C|-|AnB|-|BnC|-|AnC|+|AnBnC|

    The question is little bit tricky.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by sudeepmansh View Post
    Thanks for your answer, but I am not satisfied with the method. Can anyone please help me to solve using the formula.

    |AUBUC| = |A|+|B|+|C|-|AnB|-|BnC|-|AnC|+|AnBnC|

    The question is little bit tricky.
    Can you answer the question I asked in my post? After that it's just a matter of putting values in the above formula.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2009
    Posts
    12

    Thumbs up

    No more information has been given in the question. Its a University question paper problem.

    But one thing, the problem makes use of "among them". Is there any hidden information in that?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by sudeepmansh View Post
    No more information has been given in the question. Its a University question paper problem.

    But one thing, the problem makes use of "among them". Is there any hidden information in that?
    I was referring to my first post plz. Have a look and see if you follow that.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2009
    Posts
    12
    yes i have seen. But whats the value for AnB, BnC and AnC. No more information about that has been given in the problem.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by sudeepmansh View Post
    yes i have seen. But whats the value for AnB, BnC and AnC. No more information about that has been given in the problem.
    Thats exactly the question I had asked.
    Plz see oldguynewstudent post

    Hint: AnBnC is a subset of AnB. Correct? So what can be the min value of n(AnB)? What about others?

    This question is all about the argument above. And oldguynewstudent has given you the ans (he might not have put it formally though)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Nov 2009
    Posts
    12

    Smile

    Ok I got it. Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A tough Inclusion-Exclusion problem..or...something else
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: July 16th 2010, 06:51 PM
  2. Inclusion - Exclusion Problem
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 15th 2010, 12:06 AM
  3. Inclusion-exclusion math problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 6th 2009, 09:28 PM
  4. help with inclusion/exclusion problem
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 12th 2009, 06:41 PM
  5. Inclusion Exclusion Dice Problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 30th 2009, 06:42 AM

Search Tags


/mathhelpforum @mathhelpforum