Question

Determine the number of ways to color n objects with 3 colors if every color must be used at least once.

Answer 1

I got the answer, but my signs are switched and I don't know what is going on :/

Answer 2

I'm supposed to get 6.... I got the right formula.. it's just that the signs are causing trouble. :/ I'm not supposed to get 48.

Answer 3

@ganeshie8 save me D:

Answer 4

p.s. I am supposed to solve for that A intersection B intersection C
that's the unknown x which is the little area where all of the colors are together, but I have decided to leave it as the A intersection B intersection C as that unknown that I'm looking for.

Answer 5

3^n - (2^n+2^n+2^n) + 3

Answer 6

whattttttttttttttttttt....................

Answer 7

but wait a sec... I need to get A intersection B intersection C by itself... I do get the formula but the signs are flipped which causes me to get a bigger number than normal

Answer 8

you have n objects and 3 colors

you want to find how many ways you can have them all same color

Answer 9

the "n" objects can be all first color

the "n" objects can be all second color

the "n" objects can be all third color

Answer 10

3 ways, right ?

Answer 11

\(n(A\cap B\cap C)=3\)

Answer 12

each color has to be used once....
I have noticed that most of A B and C have one color which is red blue green so that's
1 1 1
then the intersections A and B, B and C, and A and C have two colors in common.
so that's 2^n
and then A U B U C is all of the colors of red blue green so that's the total which is 3^n

the unknown x guy is A intersection B intersection C and that is supposed to come out to 6 when n = 3 because we need at least 3 objects if we're using 3 colors

Answer 13

\(n(A\cap B\cap C)\) represents the number of ways that all the objects have same color

Answer 14

if you have 3 colors, you have 3 choices to paint all the objects with same color

Answer 15

:/

Answer 16

which is either
Red red red
green green green
blue blue blue

Answer 17

yes, those are the 3 choices

Answer 18

so how does that list become related to the 3^n - (2^n+2^n+2^n) + 3 result that I'm supposed to get :/

Answer 19

it can't be just that list that I just typed 2 minutes ago..
let's see A = red, B = blue, C = green
on most of the circle of A, B, and C there is just that one color
so absolute value of A is 1
absolute value of B is 1
absolute value of C is 1

Answer 20

sorry i have misinterpreted your venn diagram
one sec

Answer 21

sorry I was supposed to just post this question for @freckles to see because that user knows the inclusion-exclusion principle and the formula... but feel free to add some stuff too. :)

Answer 22

maybe let wait

Answer 23

Consider
\[f: \{1,2,3,4, \cdots , n\} \rightarrow \{R,B,Y\}\]
\[\text{ The number of functions we can defined } f \text{ as } \\ \text{ is } 3^n\]
\[\text{ We want only surjective functions } \\ \text{ since we want every element in the codomain to get hit at least once }\]

\[ \text{ So } 3^n \text{ is definitely too many functions... so we need to tact on} -3 \cdot 2^n\]

\[\text{ So what functions aren't surjective... functions } \\ \text{ whose elements in the codomain that don't all get hit }\]

\[\text{ So the following are problem functions: } \\ f:\{1,2,3,4, \cdots , m\} \rightarrow \{R,B\}
\text{ (since Y yellow won't get hit )}\]
\[\text{ So the following are problem functions: } \\ f:\{1,2,3,4, \cdots , m\} \rightarrow \{Y,B\}
\text{ (since R red won't get hit )}\]
\[\text{ So the following are problem functions: } \\ f:\{1,2,3,4, \cdots , m\} \rightarrow \{R,Y\}
\text{ (since B blue won't get hit )}\]
\[\text{ So we need to subtract out these functions }\]
\[\text{ But by doing that we have subtracted out too much } \]
\[\text{ We need to add back in the functions } \\ \text{ where one element gets because we have actually taking this out too many times }\]
So we need to add in:
\[f:\{1,2,3,4, \cdots m\} \rightarrow \{R\} \\ \text{ and } f:\{1,2,3,4, \cdots m\} \rightarrow \{B\} \\ \text{ and } f:\{1,2,3,4, \cdots m\} \rightarrow \{Y\} \\ \text{ so we need to tact on +3} \\ \]
\[\text{so we have the number of surjective functions from } \{1,2,3,4...,m\} \text{ to } \{R,Y,B\} \\ \text{ is } 3^{n}-3 \cdot 2^{n} +3\]

so I included all the functions from {1,2,3,...m} to {R,Y,B}
then I excluded all the functions from {1,2,3,...m} to {R,Y}, {R,B}, and {Y,B}
and then I included all the functions from {1,2,3,..m} to {R},{Y},{B}

Answer 24

oops I don't know how m become n

Answer 25

I mean n become m

Answer 26

(so replace my m's with n's please)
also n is greater than or equal to 3

Answer 27

those problem functions gave us the function where all red gets hit twice
and all yellow gets hit twice
and all blue gets hit twice
so that got too much excluded in that one step

so that is why we did the last corrective step:
so that is why we had to add in those 1+1+1 part in

Answer 28

what I'm meaning to say
for example:
is f:{1,2,3,...n} to {R,Y} and f:{1,2,3,...n} to {R,B}
...where we count the number of possible functions is 2^n and 2^n respectively
like this includes the function f:{1,2,3,...n} to {R} in both

Answer 29

does this make sense what I'm saying?

Answer 30

yes

Answer 31

cool