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Maths Problem

  • 04-02-2008 3:48pm
    #1
    Registered Users, Registered Users 2 Posts: 32


    Would appreciate any help on the mahs problem below. Asked my friends and we can't get the same answer...:confused:


    The picture shown below is to be colored in.
    (http://i254.photobucket.com/albums/hh94/katiewobbles/Teammaths98permutation.jpg)
    Five different colours are available. Each region is to be a single colour and no two adjacent regions are to be the same colour. How many different versions are possible?


Comments

  • Registered Users, Registered Users 2 Posts: 674 ✭✭✭kaki


    The way I'd look at it is that it's a sky, a sun, a road and two fields.

    Take the sun: it can be coloured in 5 different ways.
    The sky: Can be coloured in 4 ways (must not be same as sun)
    The path: Can be coloured in 4 ways (must not clash with sky)
    The left field: Can be coloured in 3 ways (mustn't clash with sky or path)
    The right field: Can be coloured in 3 ways (mustn't clash with sky or path)

    Then I think you just multiply the number of possibilites for each section together: 5x4x4x3x3=720...I think. Am totally open to correction though :o


  • Registered Users, Registered Users 2 Posts: 32 katiewobbles


    Thats EXACTLY how i did it!
    I only had doubts because it was a different answer to that of a girl who's a super genius at maths and i didn't understand her way so i had hoped that someone else could back up my version or make the other version clearer.
    Thanks btw!


  • Registered Users, Registered Users 2 Posts: 32 katiewobbles


    Problem No.2:
    A circle has both the X-axis and the Y-axis as tangents. Its centre lies on the segment from (0,10) to (15,0). Find its equation.

    Would appreciate any help:)


  • Registered Users, Registered Users 2 Posts: 32 katiewobbles


    Solved the last question on the circle, that's fine
    :):)


  • Closed Accounts Posts: 534 ✭✭✭sd123


    Problem No.2:
    A circle has both the X-axis and the Y-axis as tangents. Its centre lies on the segment from (0,10) to (15,0). Find its equation.

    Would appreciate any help:)


    Just had a quick look at this one, my maths is getting a bit rusty though.

    let centre of circle be (-h,-k), h = r = k

    equation on line passing between (15,0) and (0,10) => 2x+3y-30=0 , centre of circle, (-h,-k) is an element of this line. but we know that h=k, therefore, h = -6 and y = -6.

    eqtn of circle = (x -h)^2 + (y-k)^2 = r^2

    r= -6

    I got x^2 + y^2 +12x - 12y +36 = 0

    Some of those equations could be wrong, I haven't done maths in over 9months... hope its right


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  • Subscribers Posts: 6,408 ✭✭✭conzy


    The Circle is a horrible question :(


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