A difficult task

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    • No, there's only 5. There's only five different items that can be missing from the selected set.
      When the fake daddies are curtailed, we have failed. When their roller coaster tolerance is obliterated, their education funds are taken by Kazakhstani phishers, and their candy bars distributed between the Botswana youth gangs, we have succeeded.
      - BIG DADDY.

    • Claudio NVKP wrote:

      Joe Bartolozzi wrote:

      K.Rokossovski wrote:

      No, there's only 5. There's only five different items that can be missing from the selected set.
      4845 ways. Calculate C(20,4)
      How so, common sense doesn't tell me that.
      Welcome to the world of combinations (one of the most confusing topics in maths). You can hit Ai and see that Im right.
      Say we have 'n' items and we have to select 'k' items from them, we use C(n,k)= n!/k!.(n-k)!
      also, a sidenote n!= n*n-1*n-2*n-3.............*1
      "We can be wrong, or we can know it, but we can't do both at the same time." ~ Heisenberg

      SENIOR ELECTION MANAGER
      HEAD OF THE FPD

    • Joe Bartolozzi wrote:

      K.Rokossovski wrote:

      No, there's only 5. There's only five different items that can be missing from the selected set.
      4845 ways. Calculate C(20,4)
      OK. Lets call the object A, B, C, D, and E. Now give me just six of those 4845 ways.
      When the fake daddies are curtailed, we have failed. When their roller coaster tolerance is obliterated, their education funds are taken by Kazakhstani phishers, and their candy bars distributed between the Botswana youth gangs, we have succeeded.
      - BIG DADDY.

    • K.Rokossovski wrote:

      Joe Bartolozzi wrote:

      K.Rokossovski wrote:

      No, there's only 5. There's only five different items that can be missing from the selected set.
      4845 ways. Calculate C(20,4)
      OK. Lets call the object A, B, C, D, and E. Now give me just six of those 4845 ways.
      its 4 out of 20, we can choose seperate pairs.
      "We can be wrong, or we can know it, but we can't do both at the same time." ~ Heisenberg

      SENIOR ELECTION MANAGER
      HEAD OF THE FPD

    • I don't understand what you mean... but again, if you think there are more than 5, give me just 6 of them.
      When the fake daddies are curtailed, we have failed. When their roller coaster tolerance is obliterated, their education funds are taken by Kazakhstani phishers, and their candy bars distributed between the Botswana youth gangs, we have succeeded.
      - BIG DADDY.

    • Claudio NVKP wrote:

      K.Rokossovski wrote:

      Claudio NVKP wrote:

      5*2
      That is correct, though the actual answer should be: (5!/3!)/2!
      So 5! Is 5, 3! is 3, and 2! is 2. So it's 2(5/3)?
      5! is five faculty, or 1*2*3*4*5=120.
      3!=1*2*3=6.
      2! = 1*2=2.
      So it is (120/6)/2 = 20/2 = 10
      In normal language: you're only picking two objects; there's five ways to pick the first one, then 4 to pick the next. You're not picking the other three objects (which would be 5!), so you should divide by the ways you could have done those.

      Then you have to eliminate duplicates, because the result currently has both AB and BA in it (depending on the pick order, but that is not relevant for the result set). How many ways are there for the order to pick two objects? well, 2!


      So if you want to make it into a formula for different ways of picking x objects from a set of y items, the formula would be:
      ways = (y!/(y-x)!)/x!
      When the fake daddies are curtailed, we have failed. When their roller coaster tolerance is obliterated, their education funds are taken by Kazakhstani phishers, and their candy bars distributed between the Botswana youth gangs, we have succeeded.
      - BIG DADDY.

    • Well, you were supposed to pic 4 objects from 20
      Lets say that the objects in the set of {A,B,C,.......,T} (20)
      Fix object 'A,B,C' and make combinations around it, they are:
      A,B,C,D
      A.B,C,E
      A,B,C,F
      ......
      17 combinations made this way. Keep on doing this and obtain 4845 ways to do so.
      "We can be wrong, or we can know it, but we can't do both at the same time." ~ Heisenberg

      SENIOR ELECTION MANAGER
      HEAD OF THE FPD

    • Carking the 6th wrote:

      17, but then it would get a bit lower and lower. Wouldn’t the next be like 16, 15, 14, and you add all of that? Wouldn’t come to 4000…
      Now just fix A,B and proceed. This is not the way to do this thing actually.
      Use C(20,4) for this.
      "We can be wrong, or we can know it, but we can't do both at the same time." ~ Heisenberg

      SENIOR ELECTION MANAGER
      HEAD OF THE FPD

    • To calculate the number of ways to select 4 items from a set of 20 distinct items, you can use the combination formula, which is denoted as "n choose k" and is calculated as:
      Combination(n,k)=n!k!(n−k)!Combination(n,k)=k!(n−k)!n!
      Where:
      • n!n! denotes the factorial of nn, which is the product of all positive integers up to nn.
      • k!k! denotes the factorial of kk.
      • nn is the total number of items.
      • kk is the number of items to be selected.
      Plugging in the values, you get:
      Combination(20,4)=20!4!(20−4)!Combination(20,4)=4!(20−4)!20!
      Let's compute this:
      Combination(20,4)=20×19×18×174×3×2×1Combination(20,4)=4×3×2×120×19×18×17Combination(20,4)=4845Combination(20,4)=4845
      So, there are 4845 ways to select 4 items from a set of 20 distinct items.

      CarKing the 6th of the Abrahamic Caliphate

    • Carking the 6th wrote:

      To calculate the number of ways to select 4 items from a set of 20 distinct items, you can use the combination formula, which is denoted as "n choose k" and is calculated as:
      Combination(n,k)=n!k!(n−k)!Combination(n,k)=k!(n−k)!n!
      Where:
      • n!n! denotes the factorial of nn, which is the product of all positive integers up to nn.
      • k!k! denotes the factorial of kk.
      • nn is the total number of items.
      • kk is the number of items to be selected.
      Plugging in the values, you get:
      Combination(20,4)=20!4!(20−4)!Combination(20,4)=4!(20−4)!20!
      Let's compute this:
      Combination(20,4)=20×19×18×174×3×2×1Combination(20,4)=4×3×2×120×19×18×17Combination(20,4)=4845Combination(20,4)=4845
      So, there are 4845 ways to select 4 items from a set of 20 distinct items.
      see
      "We can be wrong, or we can know it, but we can't do both at the same time." ~ Heisenberg

      SENIOR ELECTION MANAGER
      HEAD OF THE FPD