1. Subscribersonhouse
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    14 May '14 11:281 edit
    So my doc (for real) gave me these pills to take but the instructions were a bit garbled, the gist is I start with 6 pills on day one and there are 33 pills to take. What is the regimen? For instance, you could start day one, 6, day two, 5, day 3, 4, etc, summation of 1-6, or 21. No cigar, 12 pills left.

    So what is the regimen that uses up exactly 33 pills.

    I know there are those with beautiful minds out there who will solve this before the ink dries on my monitor Joe 🙂 but put the answer in that hidden icon thing or PM me the results so other people can solve it also!
  2. Subscriberjoe shmo
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    14 May '14 22:263 edits
    Originally posted by sonhouse
    So my doc (for real) gave me these pills to take but the instructions were a bit garbled, the gist is I start with 6 pills on day one and there are 33 pills to take. What is the regimen? For instance, you could start day one, 6, day two, 5, day 3, 4, etc, summation of 1-6, or 21. No cigar, 12 pills left.

    So what is the regimen that uses up exactly 33 pil ...[text shortened]... put the answer in that hidden icon thing or PM me the results so other people can solve it also!
    I don't know about a beautiful mind...although it was a good book, but I will say this: Unless you are implying with your example that the number of pills is sequentially decreasing on each consecutive day ( which I assume you are, but just wanted to clarify) there are C(32,6) solutions.

    C(32,6) = 32!/(6!(26!)) = 32*31*30*29*28*27/(6*5*4*3*2*1) = 906,192 solutions to the problem as stated!

    Edit: I think.
  3. Subscribersonhouse
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    14 May '14 23:033 edits
    Originally posted by joe shmo
    I don't know about a beautiful mind...although it was a good book, but I will say this: Unless you are implying with your example that the number of pills is sequentially decreasing on each consecutive day ( which I assume you are, but just wanted to clarify) there are C(32,6) solutions.

    C(32,6) = 32!/(6!(26!)) = 32*31*30*29*28*27/(6*5*4*3*2*1) = 906,192 solutions to the problem as stated!

    Edit: I think.
    Where does the 32 come in? There are 33 pills and the regimen disposes of them at the last day, yes it is decreasing. I didn't know there could be THAT many solutions! Why are they factorials? why aren't they just summation kind of things?

    Can you give specific examples of what works? I know what I worked out but would be interested in other possibilities also. Another restraint on the solution is the steps have to be even, that is to say one step with X pills next step even down, even down, etc. Does that constrain the solutions?
  4. Subscriberjoe shmo
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    15 May '14 01:221 edit
    Originally posted by sonhouse
    Where does the 32 come in? There are 33 pills and the regimen disposes of them at the last day, yes it is decreasing. I didn't know there could be THAT many solutions! Why are they factorials? why aren't they just summation kind of things?

    Can you give specific examples of what works? I know what I worked out but would be interested in other possibilitie ...[text shortened]... ay one step with X pills next step even down, even down, etc. Does that constrain the solutions?
    I'll try,

    for starters lets say you had 7 pills to take in 3 days.

    1+1+1+1+1+1+1 = 7 ( you should notice there are 7 1's, and 6 +'s in the sequence )

    If you randomly pick 2 of the plus signs, and remove the rest ( I'll pick plus sign 2 and 5 for instance, removing 1,3,4,6) you get a sequence that looks like

    1...1+1...1...1+1...1 = 7

    That breaks the sum into 3 smaller summands, grouped it would look like this.

    (1...1)+(1...1...1)+(1...1) = 7, where what is inside the parenthesis are the sums

    (2)+(3)+(2) = 7, meaning that if you were to take 7 pills over 3 days, one way to do it would be:

    Day1: 2 pills
    Day2: 3 pills
    Day3: 2 pills

    But this is only one way, what if I had selected plus sign's 1 & 2 as the ones to keep, then:

    1+1+1...1...1...1...1 = 7
    (1)+(1)+(1...1...1...1...1) = 7
    (1)+(1)+(5) = 7

    Day1: 1pills
    Day2: 1 pills
    Day3: 5 pills

    In total the number of ways of selecting 2 plus signs out of 6 ( with no distinction between any plus signs) is given by:

    C(n,r) = n!/(r!(n-r)!)

    So for our example:

    n = 6 ( then number of +'s in the sequence, which is one less than the number of pills in our case)
    r = 2 ( the number of summands you wish to have minus one)

    So, C(6,2) = 6!/(2!(6-2)!) = 6*5*4!/(2*4!) = 6*5/2 = 15

    That is to say there are 15 distinct ways you can take 7 pills over 3 days.

    From here its just arithmetic to extend the idea to 33 pills over 7 days!

    And to answer your question,yes, every constraint you make weeds out some of the solutions, it gets more complicated to determine because every constraint limits the ways you can pick the +'s in the sequence, thus making the denominator of our equation larger, and the number of possible solutions smaller.
  5. Subscribersonhouse
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    15 May '14 13:35
    Originally posted by joe shmo
    I'll try,

    for starters lets say you had 7 pills to take in 3 days.

    1+1+1+1+1+1+1 = 7 ( you should notice there are 7 1's, and 6 +'s in the sequence )

    If you randomly pick 2 of the plus signs, and remove the rest ( I'll pick plus sign 2 and 5 for instance, removing 1,3,4,6) you get a sequence that looks like

    1...1+1...1...1+1...1 = 7

    That bre ...[text shortened]... hus making the denominator of our equation larger, and the number of possible solutions smaller.
    The last restraint: my regimen is 9 days. How many combo's for that?
  6. Standard memberforkedknight
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    15 May '14 19:58
    Originally posted by sonhouse
    The last restraint: my regimen is 9 days. How many combo's for that?
    There are 34c8 ways to take 33 pills in 9 days (assuming the pills are all the same)

    With the restriction that you take 6 on the first day, that leaves 28c7 = 1184040 ways to take the remaining pills in the following 8 days.
  7. Standard memberforkedknight
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    15 May '14 19:591 edit
    Originally posted by joe shmo
    I'll try,

    for starters lets say you had 7 pills to take in 3 days.

    1+1+1+1+1+1+1 = 7 ( you should notice there are 7 1's, and 6 +'s in the sequence )

    If you randomly pick 2 of the plus signs, and remove the rest ( I'll pick plus sign 2 and 5 for instance, removing 1,3,4,6) you get a sequence that looks like

    1...1+1...1...1+1...1 = 7

    That bre ...[text shortened]... hus making the denominator of our equation larger, and the number of possible solutions smaller.
    You are leaving out the cases where you take 0 pills on a given day.

    |1|1|1|1|1|1|1|

    There are 8 '|'s where you could possibly put a '+' to complete the equation for 7 pills in 3 days.

    There are 8c2 ways to take 7 pills in 3 days.
  8. Subscribersonhouse
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    16 May '14 10:333 edits
    Originally posted by forkedknight
    There are 34c8 ways to take 33 pills in 9 days (assuming the pills are all the same)

    With the restriction that you take 6 on the first day, that leaves 28c7 = 1184040 ways to take the remaining pills in the following 8 days.
    That would be 34 to the 8th power or 1.2 TRILLION ways to take those pills? Good grief!

    Here is my full regimen: Two sets of pills each day, say one set in morning, next set 12 hours later, so first day, 3,3/ that 3 times. Then minus one pill in each set of 2, so 2,2 day 4 through day 6, then 1 pill less, so day 7, 1 and 1, through day 9 which ends the treatment for a total of 33 pills.

    How many of THOSE kind of regimen's are there then?

    That is to say, regimens that follow a logical downward consumption path like this one.

    It's fun to work out math stuff like this based on real events!
  9. Standard memberforkedknight
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    16 May '14 14:11
    Originally posted by sonhouse
    That would be 34 to the 8th power or 1.2 TRILLION ways to take those pills? Good grief!

    Here is my full regimen: Two sets of pills each day, say one set in morning, next set 12 hours later, so first day, 3,3/ that 3 times. Then minus one pill in each set of 2, so 2,2 day 4 through day 6, then 1 pill less, so day 7, 1 and 1, through day 9 which ends the t ...[text shortened]... nsumption path like this one.

    It's fun to work out math stuff like this based on real events!
    34c8 = 34!/(8!*26!) = 18156204

    Not quite a trillion, but 18 million ways.
  10. Standard memberforkedknight
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    16 May '14 18:391 edit
    Originally posted by sonhouse

    Here is my full regimen: Two sets of pills each day, say one set in morning, next set 12 hours later, so first day, 3,3/ that 3 times. Then minus one pill in each set of 2, so 2,2 day 4 through day 6, then 1 pill less, so day 7, 1 and 1, through day 9 which ends the treatment for a total of 33 pills.
    Also, I count 36 pills in your regimen
    6+6+6+4+4+4+2+2+2 = 36

    You may want to talk to your pharmacist if he only gave you 33 pills 🙂
  11. Subscribersonhouse
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    16 May '14 22:37
    Originally posted by forkedknight
    Also, I count 36 pills in your regimen
    6+6+6+4+4+4+2+2+2 = 36

    You may want to talk to your pharmacist if he only gave you 33 pills 🙂
    No, that last jump loses two pills. It is a steady one pill loss at a time so the last jump is 1,1,1 or 33.
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    06 Jun '14 01:22
    6 pills every day on days 1-5. On day 6 take the last 3.
  13. Subscribersonhouse
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    06 Jun '14 16:33
    Originally posted by Flower04
    6 pills every day on days 1-5. On day 6 take the last 3.
    That regime doesn't titrate down like the doctor ordered.
  14. SubscriberAThousandYoung
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    07 Jun '14 22:34
    655443321
  15. Subscribersonhouse
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    04 Jul '14 16:00
    Originally posted by AThousandYoung
    655443321
    That's pretty much what the doctor ordered.
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