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## Precalculus

### Course: Precalculus>Unit 8

Lesson 3: Permutations

# 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.

## Want to join the conversation?

• What if the colours CAN repeat? How would the process/equation change? •   6 x 6 x 6 x 6
Think about it like this: Since the colours can repeat and there are 6 alternatives, there will be 6 possible colours for every slot.
• 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? •  Lets suppose you have just 2 bits. How many numbers can you represent?

``0  0  =  00  1  =  11  0  =  21  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 = 00 0 1 = 10 1 0 = 20 1 1 = 31 0 0 = 41 0 1 = 51 1 0 = 61 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.
• in how many ways can the first, second, and third placers be chosen from a group of 8 contestants? • 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.
• can anyone tell me why the factorial of 0 is 1? • And what if BRYG is considered the same as GRYB? What numbers are input into the spaces then, for calculating the answer? • 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!
• 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. :( • in how many ways can 30 people be divided into 15 couples • 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
• 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....   