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Math Help - Mixed bag part 2: pemutations, combinations and determinants

  1. #1
    Member MathBlaster47's Avatar
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    Mixed bag part 2: pemutations, combinations and determinants

    I just want to make sure I have my method down properly, as usual, so here goes another round of questions.

    1:
    There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if equal numbers of men and women are to be on the committee?

    Method:

    There are  _{10}W_3 ways to select 3 women for the committee and  _8M_3 ways to select 3 men for the committee. Therefore, there are _{10}W_3\cdot_8M_3 different committees possible, so _{10}W_3 \cdot _8M_3 \rightarrow \frac{10!}{(10-3)!} \cdot \frac{8!}{(8-3)!}\rightarrow\frac{10!}{7!}\cdot\frac{8!}{5!}\ri  ghtarrow \frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3  \cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot  1}\cdot\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2  \cdot1}{5\cdot4\cdot3\cdot2\cdot1}\rightarrow\frac  {3628800}{5040}\cdot\frac{40320}{120}=241920
    Thus there are 241920 possible committees.

    2:
    Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white.

    Method:
    There are 16 marbles in all so there are _{16}M_2\rightarrow\frac{16!}{14!}\rightarrow \frac{2.09227899 \cdot 10^{13}}{87178291200}=240ways of picking 2 marbles.
    This is where I get fuzzy, I'm not too sure what to do next--should I multiply out all the permutations for all the colors and then work from there?
    Or should I just focus on the white marbles?

    3:
    If two rows of a determinant are interchanged, what is true of the resulting determinant?

    Method:
    Sadly for me, I have little clue of what to do. I could tell you why if one row of a determinant is 0 the whole things value is 0, but I don't know where to start on this one.

    Thank you
    Last edited by MathBlaster47; March 1st 2010 at 07:33 AM.
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  2. #2
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    Hello, MathBlaster47!

    The first two are Combination problems, not Permutations.


    1) There are 10 women and 8 men in a club.
    How many different committees of 6 people can be selected from the group
    if 3 women and 3 women are to be on the committee?
    There are: . _{10}C_3 \:=\:{10\choose3} \:=\:\frac{10!}{3!\,7!} \:=\:120 ways to choose 3 women.

    There are: . _8C_3 \:=\:{8\choose3} \:=\:\frac{8!}{3!\,5!} \:=\:56 ways to choose 3 men.

    Therefore, there are: . 120\cdot56 \:=\:6720 committees with 3 women and 3 men.





    2) Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles.
    Find the probability of both marbles being white.

    There are: . _{16}C_2 \:=\:{16\choose2} \:=\:\frac{16!}{2!\,14!} \:=\:120 possible two-marbles samples.

    There are: . _6C_2 \:=\:{6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\: 15 ways to get two white marbles.

    Therefore: . P(\text{two whites}) \;=\;\frac{15}{120} \;=\;\frac{1}{8}




    3) If two rows of a determinant are interchanged,
    what is true of the resulting determinant?
    It depends on which two rows are interchanged.

    If the two rows are an odd number of rows apart,
    . . the sign of the determinant is changed.

    If the two rows are an even number of rows apart,
    . . the value of the determinant is unchanged.

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  3. #3
    Junior Member StonerPenguin's Avatar
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    Wow! On exam 11 and 12! Almost there~! Chapter 16 was a real mo' fo' for me for whatever reason.. Anywho, thanks for asking these questions, I was a little unsure too, but my answers were the same as Soroban's *insert nerdy math gloating here* (rofl)

    But for the last question;
    If two rows of a determinant are interchanged, what is true of the resulting determinant?

    I wouldn't put the "If the two rows are an even number of rows apart, the value of the determinant is unchanged." part Soroban's answer (no offense), because our textbook simply doesn't cover it. That chapter really only discussed 3 x 3 determinants. Look at chapter 16-8 Properties of Determinants (page 798) Property 3
    "If two rows (or two columns) of a determinant are interchanged, then the resulting determinant is the opposite of the original determinant."

    Even though that's not true for all determinants as Soroban pointed out, that's how they want you to answer D':
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  4. #4
    Member MathBlaster47's Avatar
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    Quote Originally Posted by StonerPenguin View Post
    Wow! On exam 11 and 12! Almost there~! Chapter 16 was a real mo' fo' for me for whatever reason.. Anywho, thanks for asking these questions, I was a little unsure too, but my answers were the same as Soroban's *insert nerdy math gloating here* (rofl)

    But for the last question;
    If two rows of a determinant are interchanged, what is true of the resulting determinant?

    I wouldn't put the "If the two rows are an even number of rows apart, the value of the determinant is unchanged." part Soroban's answer (no offense), because our textbook simply doesn't cover it. That chapter really only discussed 3 x 3 determinants. Look at chapter 16-8 Properties of Determinants (page 798) Property 3
    "If two rows (or two columns) of a determinant are interchanged, then the resulting determinant is the opposite of the original determinant."

    Even though that's not true for all determinants as Soroban pointed out, that's how they want you to answer D':
    Ah, thank you! I don't mind gloating--you did earn it!
    I however, did not....alas those permutations and combination problems are not my strong suit, statistics, and much of trig I can do no problem.
    Thanks for catching my oversight!

    Quote Originally Posted by Soroban View Post
    Hello, MathBlaster47!

    The first two are Combination problems, not Permutations.


    There are: . _{10}C_3 \:=\:{10\choose3} \:=\:\frac{10!}{3!\,7!} \:=\:120 ways to choose 3 women.

    There are: . _8C_3 \:=\:{8\choose3} \:=\:\frac{8!}{3!\,5!} \:=\:56 ways to choose 3 men.

    Therefore, there are: . 120\cdot56 \:=\:6720 committees with 3 women and 3 men.






    There are: . _{16}C_2 \:=\:{16\choose2} \:=\:\frac{16!}{2!\,14!} \:=\:120 possible two-marbles samples.

    There are: . _6C_2 \:=\:{6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\: 15 ways to get two white marbles.

    Therefore: . P(\text{two whites}) \;=\;\frac{15}{120} \;=\;\frac{1}{8}




    It depends on which two rows are interchanged.

    If the two rows are an odd number of rows apart,
    . . the sign of the determinant is changed.

    If the two rows are an even number of rows apart,
    . . the value of the determinant is unchanged.

    Thank you!
    I realize where I went wrong....dang things looking so similar.....
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  5. #5
    Junior Member StonerPenguin's Avatar
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    Eh, that's cool. Determinants and matrices really aren't my suit so I know how it is. And the important thing is that you realize your mistakes, which is the key to improvement. And the users on here are really good at showing their process and helping, which is why I this forum lawl
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