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Posers and Puzzles

Posers and Puzzles

  1. 10 May '14 16:46 / 1 edit
    My wallet $7 lighter, I walked out of the store that sells used books with a book of math puzzles. I'll give one of the easiest ones from the book, plus a medium-difficult one. Maybe later, a difficult one.

    I.
    A wooden cube has edges of length 3 meters. Square holes of side one meter, centered in each face, are cut through to the opposite face. The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside is how many square meters?



    II.
    A box contains 2 pennies, 4 nickels and 6 dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chose. What is the probability that the value of the coins drawn is at least 50 cents?

    [For those outside the USA, penny = 1 cent, nickel = 5 cents, dime = 10 cents.]
  2. Subscriber joe shmo On Vacation
    Strange Egg
    10 May '14 18:13
    Originally posted by Paul Dirac II
    My wallet $7 lighter, I walked out of the store that sells used books with a book of math puzzles. I'll give one of the easiest ones from the book, plus a medium-difficult one. Maybe later, a difficult one.

    I.
    A wooden cube has edges of length 3 meters. Square holes of side one meter, centered in each face, are cut through to the opposite face. The ...[text shortened]... ast 50 cents?

    [For those outside the USA, penny = 1 cent, nickel = 5 cents, dime = 10 cents.]
    I. 72 m²

    A_surface = 6*3² - 6*1² + 6*4*1 [m²]

    = 6(9 -1 +4)
    = 6(12)
    =72
  3. 10 May '14 18:32 / 1 edit
    Originally posted by joe shmo
    I. 72 m²
    Give that man Joe a gold star.


    Oh, and I've got to ask you how you make a superscript here, as in the square on the meters.
  4. Subscriber joe shmo On Vacation
    Strange Egg
    10 May '14 19:50 / 1 edit
    Originally posted by Paul Dirac II
    Give that man Joe a gold star.


    Oh, and I've got to ask you how you make a superscript here, as in the square on the meters.
    Thanks

    The superscript is not a general one , you can only make superscript 2. They are Alt Codes.

    http://www.alt-codes.net/

    However, don't get too excited...many of these do not decode in this forum, but some do.
  5. Subscriber joe shmo On Vacation
    Strange Egg
    10 May '14 23:44 / 1 edit
    Originally posted by joe shmo
    Thanks

    The superscript is not a general one , you can only make superscript 2. They are Alt Codes.

    http://www.alt-codes.net/

    However, don't get too excited...many of these do not decode in this forum, but some do.
    II. P = 4/C(12,6)

    C(12,6) = 12!/(6!*(12-6)!)

    C(12,6) = 924

    P = 4/924
  6. 10 May '14 23:50 / 1 edit
    Originally posted by joe shmo
    II. P = 4/C(12,6)

    C(12,6) = 12!/(6!*(12-6)!)

    C(12,6) = 924

    P = 4/924
    The book gives the answer as being a substantially higher probability than that. How do you justify the 4 in the numerator of your first line?
  7. Subscriber joe shmo On Vacation
    Strange Egg
    11 May '14 00:05 / 1 edit
    Originally posted by Paul Dirac II
    The book gives the answer as being a substantially higher probability than that. How do you justify the 4 in the numerator of your first line?
    I suppose Iv'e overlooked something. I assumed there were only 4 favorable cases where the sum of the 6 coins is >= 50 cents

    Namely:

    10,10,10,10,10,10
    ....................,5
    ....................,1
    ..................5,5

    I'm admittedly weak in combinatorics.
  8. 11 May '14 00:53 / 2 edits
    It might help to briefly consider an alternative problem where the box contains two pennies, four nickels, and a million dimes. Intuitively, the probability of totaling at least 50 cents when pulling six coins out randomly from such a box is almost 100%.

    Another thing that might help in the case of the box of 2 + 4 + 6 coins is to imagine some sneaky person has penciled labels on the coins. For example, one of the dimes might have 'A' written on it, another dime might have 'B' written on it, and so on.
  9. 11 May '14 18:17
    A hint: the book presents the answer as a fraction. Joe's denominator matches the book's denominator.
  10. 12 May '14 01:25 / 4 edits
    I get:
    12 coins
    number of ways of drawing 6 coins = 12*11*10*9*8*7

    there are 4 possible solutions:
    6d
    1n 5d
    1p 5d
    2n 4d

    first, consider how many different combinations with 'labelled coins' there are

    With 6d we have used all our dimes, so only one combination (permutations come later)
    with 1n 5d,we can have 1 out of 4 nickels and miss out any one dime, so 4*6 combinations
    similarly for 1p 5d, 2*6 combinations

    2n 4d is more complex
    2n can be selected in 4*3/2 = 6 ways
    2d can be selected in 6*5/2 = 15 ways
    giving 6*15 combinations

    so the total combinations giving a solution is:
    1 + 4*6 + 2*6 + 15*6 = 1 + 21*6 = 127

    each combination has 6*5*4*3*2 permutations
    giving a total 127*6*5*4*3*2 solution permutations

    So the probability of getting 50 is:

    127*6*5*4*3*2/(12*11*10*9*8*7)

    = 127/(11*12*7)

    = 127/924
  11. 12 May '14 01:42 / 2 edits
    Originally posted by iamatiger

    = 127/924
    The book agrees with your 127/924.
  12. 12 May '14 01:44 / 1 edit
    Originally posted by Paul Dirac II
    The book agrees with your 127/924.
    Great!
  13. 12 May '14 07:52
    Its quite tricky to calculate the average total got with 6 random coins.
  14. 12 May '14 20:53
    Anyone want to work out average total with 6 random coins?
  15. Standard member forkedknight
    Defend the Universe
    12 May '14 23:15
    Originally posted by iamatiger
    Anyone want to work out average total with 6 random coins?
    Intuitively, it should be 41 cents. You're taking half the coins, so on average, you should have half your pennies, half your nickles, and half your dimes.