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Ways to arrange colors

Thinking about how many ways you can pick four colors from a group of 6. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • blobby green style avatar for user maria.cm7
    What if the colours CAN repeat? How would the process/equation change?
    (26 votes)
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  • purple pi purple style avatar for user Darko
    Am not much into binary numbers and such but am wondering... Why can we store 256 different values into 8 bits and why does 8 bit translate into a decimal value of 4 decimal spaces? I know that an 8 bit color has 256 levels of black and white. And I can get any color when I mix 3 basic colors, Red, Green and Blue and using combinatronics I get 256^3 since these 3 colors can have same value or same level of shade between black and white. But why does 8 bit translate to 256 shades?
    (10 votes)
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    • leaf blue style avatar for user Stefen
      Lets suppose you have just 2 bits. How many numbers can you represent?

      0  0  =  0
      0 1 = 1
      1 0 = 2
      1 1 = 3

      So with just 2 bits we can represent 4 numbers. Notice that 2^2 = 4
      Let's add another bit, now how many numbers can we represent?
      0 0 0 = 0
      0 0 1 = 1
      0 1 0 = 2
      0 1 1 = 3
      1 0 0 = 4
      1 0 1 = 5
      1 1 0 = 6
      1 1 1 = 7

      So with three bits we can represent 8 numbers, 0 to 7. Notice that 2^3 = 8.
      That is the pattern. So, with 8 bits we can represent 2^8=256 different numbers or levels.
      (27 votes)
  • blobby green style avatar for user Ems Gustilo Piolo
    in how many ways can the first, second, and third placers be chosen from a group of 8 contestants?
    (6 votes)
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    • male robot hal style avatar for user Lizard Wizard Ninja
      Sorry i'm replying to you 8 years later, but it's better than nothing.

      If you watch the Permutation formula video, you see that if you don't have enough spots for every position, you take the places, which there are 3 in this case, and then start from 8 and count down 3, if it was 4 you would count down 4, etc. So 8,7, and 6. So you multiply 8 7 and 6. And you get 336 ways that first second and third placers be chosen from 8 people.
      (5 votes)
  • male robot johnny style avatar for user paradox ujjawal
    can anyone tell me why the factorial of 0 is 1?
    (6 votes)
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  • leaf red style avatar for user Raymando
    And what if BRYG is considered the same as GRYB? What numbers are input into the spaces then, for calculating the answer?
    (5 votes)
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    • blobby green style avatar for user robshowsides
      Good question! He mentions that because he considers them different, he is finding permutations (P). When you consider BRYG and GRYB the same, that is called finding the number of combinations (C). You can see that there will (almost) always be fewer combinations than permutations, since lots of permutations will only count as a single combination. If you are choosing r things out a collection of n things (in this case we are choosing 4 colors out of a total of 6 colors, so r = 4 and n = 6), then C(n, r) = P(n, r) / r!, where the "!" means "factorial" (3! = 3*2*1 = 6, 5! = 5*4*3*2*1 = 120, etc.). So in this problem, the number of combinations would equal the number of permutations divided by 4! = 4*3*2*1 = 24. So if BRYG, GRYB, RGBY, etc., were considered the same, then instead of 360, the number of possibilities would be 360/24 = 15. Not very many!
      (7 votes)
  • blobby green style avatar for user Emma Davenport
    this is a question about homework but this is not the real question: If the numbers 1-8 are used to make 3-digit numbers and they do not reapt , how many 3-digit numbers can be made? What is the formula? I don't get this. :(
    (5 votes)
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  • blobby green style avatar for user venkateswararao1111
    in how many ways can 30 people be divided into 15 couples
    (5 votes)
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    • blobby green style avatar for user Michelle Hein
      If you have 2 people and 1 couple, then there is only 1 option.

      If you have 4 people and 2 couples, then the first person will select from 3 people. 2 people are left after they split off, but there is only 1 way to combine 2 people so we get 3x1.

      If you have 6 people and 3 couples, then the first person chooses from 5 people and that leaves 4 people to make 2 couples. From above this is 3x1. So all together this makes 5x3x1
      You can keep going like this and you'll see that you end up with a factorial of odd numbers up through 29. To get this, you can divide 30! by the factorial of even numbers through 30. This looks like (2^15)x15! because (2x2x2...2)(1x2x3....15) = (2x4x6...30). So all together it would look like
      30! / [(2^15) x 15!] = 6.19 x 10^15
      (4 votes)
  • scuttlebug green style avatar for user Tasya Adzkiya
    Well, I think it matters because the "The codemaker gives hints about whether the colors are correct and IN THE RIGHT POSITION." shows us that we can't mix it up (the order matters). Because if you are trying to decipher a code, the arrangement of the code has to be in order. For example, B L U E is different than U L E B although the letters are the same. That is a hint for us to use permutation instead of combination because order matters in permutation.
    (4 votes)
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    • blobby green style avatar for user ersepsi
      To be honest, I still think Sal is right, but I understand your thoughts. The question is basically the paragraph but shortened. In the paragraph, it's mentioned that the codemaker says whether the color is CORRECT or NOT, therefore we know the colors CANNOT be repeated.

      If the codemaker doesn't give any hints, then the color COULD be repeated. But that would obviously be included in the question. That's why I agree with Sal that the paragraph doesn't really matter.
      (4 votes)
  • leaf green style avatar for user Ken Bonifay
    how would you solve the problem:

    "How many possible options are there to place 12 golfers on three teams of four?
    (4 votes)
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    • leaf green style avatar for user kubleeka
      Imagine you have the 12 golfers lined up (labelled as 'g') with two separators to group them into teams.
      gggg|gggg|gggg

      If you leave the separators in place, you can reorder the golfers in 12! ways, which will divide them into teams in every possible way with some overcounting.

      Each team can be ordered in 4! ways, so there are (4!)³ ways to reorder the golfers within their teams. For each of those orderings, we can also permute the teams themselves in 3!=6 ways. That is, our 12! figure counts each team (4!)³·6 ways, so the correct number is
      12!/((4!)³·6)=5775.
      (4 votes)
  • leafers ultimate style avatar for user Shreyas Pai
    I have an interesting problem on combinatorics for football (soccer) lovers...
    Try to work out how many games are played in one season of the Barclays Premier League... (in England)
    [For those of you who aren't football lovers, here are some clues:
    1.There are 20 teams in the Premier League.
    2.Each team plays every other team twice, once at home and once away.]
    P.S. I was watching a match and suddenly wondered how many matches are played in total - that's one of the reasons why I wanted to learn about this topic....
    (3 votes)
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Video transcript

In one game, a code made using different colors is created by one player, the codemaker, and the other player, the codebreaker, tries to guess the code. The codemaker gives hints about whether the colors are correct and in the right position. All right. The possible colors are blue-- let me underline these in the actual colors-- blue, yellow, white, red, orange and green. Green is already written in green, but I'll underline it in green again. And green. How many 4-color codes can be made if the colors cannot be repeated? To some degree, this whole paragraph in the beginning doesn't even matter. If we're just choosing from-- let's see, we're choosing from-- how many colors are there? There's 1, 2, 3, 4, 5, 6 colors, and we're going to pick 4 of them. How many 4-color codes can be made if the colors cannot be repeated? And since these are codes, we're going to assume that blue, red, yellow and green, that this-- that that is different than green, red, yellow and blue. We're going to assume that these are not the same code. Even though we've picked the same 4 colors, we're going to assume that these are 2 different codes, and that makes sense because we're dealing with codes. So these are different codes. So this would count as 2 different codes right here, even though we've picked the same actual colors. The same 4 colors, we've picked them in different orders. Now, with that out of the way, let's think about how many different ways we can pick 4 colors. So let's say we have 4 slots here. 1 slot, 2 slot, 3 slot and 4 slots. And at first, we care only about, how many ways can we pick a color for that slot right there, that first slot? We haven't picked any colors yet. Well, we have 6 possible colors, 1, 2, 3, 4, 5, 6. So there's going to be 6 different possibilities for this slot right there. So let's put a 6 right there. Now, they told us that the colors cannot be repeated, so whatever color is in this slot, we're going to take it out of the possible colors. So now that we've taken that color out, how many possibilities are when we go to this slot, when we go to the next slot? How many possibilities when we go to the next slot right here? Well, we took 1 of the 6 out for the first slot, so there's only 5 possibilities here. And by the same logic when we go to the third slot, we've used up 2 of the slots-- 2 of the colors already, so there would only 4 possible colors left. And then for the last slot, we would've used up 3 of the colors, so there's only 3 possibilities left. So if we think about all of the possibilities, all of the permutations-- and permutations are when you think about all the possibilities and you do care about order; where you say that this is different than this-- this is a different permutation than this. So all of the different permutations here, when you pick 4 colors out of a possible of 6 colors, it's going to be 6 possibilities for the first 1, times 5 for the second bucket, times 4 for the third or the third bucket of the third position, times 3. So 6 times 5 is 30, times 4 is times 3. So 30 times 12. So this is 30 times 12, which is equal to their 360 possible 4-color codes.