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Thread: Combination

  1. #1
    Junior Member HeroWise's Avatar
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    Combination

    THis is from the 2015 perms thread
    A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
    In how many ways can he place the tigers if no cage to be left empty?


    I got

    4! x (6C2 x 4C2 x 2C1 x 1C1) = 4320

    Because: First Cage: 6C2
    Second Cage: 4C2
    Third Cage: 2C1
    Fourth Cage: 1C1
    4! ways to arrange these

    Some people got 2160 which is 4320/2
    some also got 1080 which is 4320/4

    IM confused to what is the right answer. If so why am I wrong

  2. #2
    617 pages fan96's Avatar
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    Re: Combination

    Are these four cages distinct or identical?

    Are the tigers distinct or identical?
    Last edited by fan96; 6 May 2018 at 1:35 PM.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    617 pages fan96's Avatar
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    Re: Combination

    Assuming the tigers and cages are identical and arranged in a straight line:

    If no cage is to be empty and each cage can fit a maximum of two tigers, then there must be either one or two tigers in each cage.

    Also, there must be exactly 6 tigers.

    The only way to make 6 from adding 1 or 2, four times, is in any order (that sentence is a bit botched but hopefully you get what I mean).

    So assuming the cages are identical and they're not arranged in a circle, the problem can be reduced to "How many ways can you arrange two 2's and two 1's?"



    Another way to solve the problem is to consider each case by making something similar to a tree diagram.



    If the tigers are distinct, multiply the answer by .

    If the cages are distinct, multiply the answer by .

    Your answer should be correct if the tigers are distinct but the cages are not ().
    Last edited by fan96; 6 May 2018 at 1:43 PM.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

  4. #4
    -insert title here- Paradoxica's Avatar
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    Re: Combination

    Quote Originally Posted by HeroWise View Post
    THis is from the 2015 perms thread
    A zoo keeper wants to place 6 tigers in 4 cages, each can fit a maximum of two tigers.
    In how many ways can he place the tigers if no cage to be left empty?


    I got

    4! x (6C2 x 4C2 x 2C1 x 1C1) = 4320

    Because: First Cage: 6C2
    Second Cage: 4C2
    Third Cage: 2C1
    Fourth Cage: 1C1
    4! ways to arrange these

    Some people got 2160 which is 4320/2
    some also got 1080 which is 4320/4

    IM confused to what is the right answer. If so why am I wrong
    Have you tried the Stars and Bars method?
    KAIO7 likes this.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  5. #5
    Junior Member HeroWise's Avatar
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    Re: Combination

    No I have not

    I have heard of it and have seen @fan96 post something about it before,

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    Re: Combination

    https://en.wikipedia.org/wiki/Stars_...combinatorics)

    Basically if you want to place objects in boxes you can model it using something like



    Where the stars are the objects and the bars represent the dividers between boxes. So in the above case you would be placing 6 objects in 3 boxes.

    By using the formula for permutations with repetition you can count the number of ways to distribute 6 objects across 3 boxes with no restriction:

    ,

    since there are 6 stars and 2 bars.

    It's pretty useful to know but for this question, since the numbers are small and there are quite a few restrictions (no more than 2 tigers in a cage, no cage empty), I don't think it's as useful here. You'd have to subtract the cases where two dividers are together (representing an empty box) and when there are more than two stars next to each other (i.e. more than two lions in a cage).
    Last edited by fan96; 6 May 2018 at 4:26 PM.
    KAIO7 likes this.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

  7. #7
    Junior Member HeroWise's Avatar
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    Re: Combination

    Can you give me a good question and can you please work it out with this method?

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    Rambling Spirit
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    Re: Combination

    Quote Originally Posted by HeroWise View Post
    Can you give me a good question and can you please work it out with this method?




    fan96 likes this.

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    Re: Combination

    Here's an easier question from one of my school's 4U Task 1 papers:

    In how many ways can nine identical rings be placed on the four fingers of the left hand, excluding the thumb, if:
    • there are no restrictions?
    • there must be at least one ring on each finger?
    • there are no restrictions and the rings are all different?


    Quote Originally Posted by InteGrand View Post


    Is that a past HSC question?
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

  10. #10
    Junior Member HeroWise's Avatar
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    Re: Combination

    fan when there are no restrictions its 220 right?

    I used the stars and bars method for it

    9 stars 3 bars

    12C3

    for ii) it is 220 - cases with a finger or till 4 having no rings.

    but how do i do it?

  11. #11
    617 pages fan96's Avatar
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    Re: Combination

    Quote Originally Posted by HeroWise View Post
    fan when there are no restrictions its 220 right?

    I used the stars and bars method for it

    9 stars 3 bars

    12C3

    for ii) it is 220 - cases with a finger or till 4 having no rings.

    but how do i do it?
    Yep it's 220.

    For the second question, the marker's answer sets up the rings as such:

    R_R_R_R_R_R_R_R_R

    Where the eight "_" characters represent possible spaces for a divider, of which there must be three. Thus the answer is simply .
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

  12. #12
    Rambling Spirit
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    Re: Combination

    Quote Originally Posted by fan96 View Post
    Here's an easier question from one of my school's 4U Task 1 papers:

    In how many ways can nine identical rings be placed on the four fingers of the left hand, excluding the thumb, if:
    • there are no restrictions?
    • there must be at least one ring on each finger?
    • there are no restrictions and the rings are all different?




    Is that a past HSC question?
    I doubt it. If it is, it would be from a long time ago and would be coincidence (I didn't pick that question out of a past HSC paper). I included it since it's a nice application of Stars and Bars.

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