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# counting problem, help

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35
9

2 different questions: a) Violet has six beads that she wants to assemble into a bracelet. Two of the beads have the same color, and the rest have different colors. How many different ways can Violet assemble her bracelet? (Two bracelets are considered identical if one can be rotated and/or reflected to obtain the other.)

b) Jay owns a fleet of  shiny cars.
Each car has either standard transmission or automatic transmission
Each car is either black, white, or red
Each car is either a Toyota, Mercedes-Benz, Chevrolet, or Pinto
No two cars are the same for each of the three categories. (For example, no two cars have standard transmission, are white, and Toyotas.)
Two of the cars are chosen at random. What is the probability that the two cars differ in all three categories? (For example, a black Chevrolet with standard transmission is different in all three categories from a red Pinto with automatic transmission, but not from a black Mercedes-Benz with automatic transmission.)

Jan 7, 2020
edited by Guest  Jan 7, 2020

#1
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There are 6*9 = 45 ways of arranging the beads.

Jan 7, 2020
#2
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Why is it 6*9? I got 6!/2! because there are 6! total possibilities dividing 2! because there are two of the same, but it was incorrect.

Also 6*9 is 54

Guest Jan 7, 2020
#3
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where did the 9 come from?

Guest Jan 7, 2020
#4
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We can place one of the unique beads, and call it reference number 1.  Then the possible places of the two same colored beads are at 2 and 6, 3 or 5 and 4, so three ways.  The rest of the 4 charms can be placed in 4! ways.  So the number of arrangements is 3*4! = 72.

Jan 7, 2020
#5
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but doesn't that only work when one of the unique beads are at place 1? what about when one of the same colored beads are at place 1?

Guest Jan 7, 2020
#9
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a) Violet has six beads that she wants to assemble into a bracelet. Two of the beads have the same color, and the rest have different colors. How many different ways can Violet assemble her bracelet? (Two bracelets are considered identical if one can be rotated and/or reflected to obtain the other.)

Without the reflection it is straight forward.

5!/2 = 60

With the reflection I am not at all sure.

Since one of the beds must  stay fixed,

Every permutation would have exactly one reflection permutation so I think maybe you divide by 2

Which would be 30

Can you please let us know when you have the answer as we can all learn that way.

Melody  Jan 8, 2020
edited by Melody  Jan 8, 2020
#6
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i have no idea how to approach b), i know it probably includes a lot of choosing but i don't know if it's multiplying/adding and i don't know how to figure out which ones differ in all three categories. even a hint would be good. thank you!!

Jan 7, 2020
#7
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There are 24 "different" cars:

(S,B,M)
(S,B,T)
(S,B,C)
(S,B,P)
(S,W,M)
(S,W,T)
(S,W,C)
(S,W,P)
(S,R,M)
(S,R,T)
(S,R,C)
(S,R,P)
(A,B,M)
(A,B,T)
(A,B,C)
(A,B,P)
(A,W,M)
(A,W,T)
(A,W,C)
(A,W,P)
(A,R,M)
(A,R,T)
(A,R,C)
(A,R,P)

Notice that there are 4 ways to pair the Toyotas with a "different" Mercedes-Benz, "different" Chevrolet or "different" Pinto.  So, 4 + 4 + 4 = 12

And there are 4 ways  to pair the Mercedes-Benz with a "different" Chevrolet or "different'' Pinto. {We've already paired each with a "different" Toyota.} So 4 + 4 = 8

And since we've already paired the Chevrolet with a "different" Toyota or Mercedes-Benz, we have 4 ways that we can pair them with a "different" Pinto.

So there are 12 + 8 + 4 = 24 "different" pairings

And the total number of possible pairs made by selecting any 2 cars from 24  = C(24,2) = 276

So the probability is 24 / 276 = 2/23.

Jan 7, 2020
#8
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