Results 1 to 5 of 5

Math Help - Combinations

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    77

    Combinations

    The questions:

    1) There are 8 free rooms in a hostel. How many ways can 4 students be divided into these rooms?

    2) A student must choose 5 question from a 12 question exam paper. If he chooses at least 3 from the first 6, calculate the number of ways he can choose.

    My answers:

    1) Since order does not matter, we calculate as such: \left(^{8}\mathcal{P}_5\right) which will give us 5040. Is this right or am I doing something wrong?

    2)Choosing at least 3 from the first 6 it will give 12X11X10 = 1320. From here, I am stuck assuming I was right upto here in the first place.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,788
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by Keep View Post
    The questions:
    1) There are 8 free rooms in a hostel. How many ways can 4 students be divided into these rooms?
    2) A student must choose 5 question from a 12 question exam paper. If he chooses at least 3 from the first 6, calculate the number of ways he can choose.
    My answers:
    1) Since order does not matter, we calculate as such: \left(^{8}\mathcal{P}_5\right) which will give us 5040. Is this right or am I doing something wrong?
    2)Choosing at least 3 from the first 6 it will give 12X11X10 = 1320. From here, I am stuck assuming I was right upto here in the first place.
    In #1 order does make a difference because presumably the rooms are numbered.
    So it is a permutation and that the notation you used but the wrong answer.
    \left(^{8}\mathcal{P}_4\right)=(8)(7)(6)(5)=1680

    For the second question it is a combination: \left(^{6}\mathcal{C}_3\right)\cdot\left(^{9}\math  cal{C}_2\right) .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2009
    Posts
    77
    Quote Originally Posted by Plato View Post
    In #1 order does make a difference because presumably the rooms are numbered.
    So it is a permutation and that the notation you used but the wrong answer.
    \left(^{8}\mathcal{P}_4\right)=(8)(7)(6)(5)=1680

    For the second question it is a combination: \left(^{6}\mathcal{C}_3\right)\cdot\left(^{9}\math  cal{C}_2\right) .
    Thank you Plato. In number one, the answer will be 1680 if we presume that the rooms are numbered and hence that order matters. But what if we assume that the rooms are not numbered and there is no order. Then it will be \left(^{8}\mathcal{C}_4\right) which is equal to 56, right?

    PS:Sorry I meant to write C, not P before.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,788
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by Keep View Post
    But what if we assume that the rooms are not numbered and there is no order. Then it will be \color{red}\left(^{8}\mathcal{C}_4\right) which is equal to 5040, right?
    No indeed. \left(^{8}\mathcal{C}_4\right)=\frac{8!}{(4!)(4!)}  =70
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2009
    Posts
    77
    Quote Originally Posted by Plato View Post
    No indeed. \left(^{8}\mathcal{C}_4\right)=\frac{8!}{(4!)(4!)}  =70

    Sorry it is all my mistakes. Nevermind. I get it now. I thought I wrote 5 rooms and then did not calculate properly.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations in a set
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 9th 2010, 06:19 AM
  2. How many combinations are possible?
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: July 23rd 2009, 07:53 PM
  3. Combinations
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 5th 2008, 08:28 AM
  4. How many combinations..?
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 2nd 2008, 10:34 AM
  5. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 27th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum