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A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals

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A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals  [#permalink]

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New post Updated on: 13 Aug 2018, 06:45
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Question Stats:

74% (00:52) correct 26% (00:53) wrong based on 117 sessions

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Learn when to “Add” and “Multiply” in Permutation & Combination questions- Exercise Question #2

A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals are made by combining any \(2\) of these \(4\) colors. How many unique signals can the traffic signal show if the order of the colors defines unique cases?


A) 3
B) 6
C) 8
D) 12
E) 24

Learn to use the Keyword Approach in Solving PnC question from the following article:

Learn when to “Add” and “Multiply” in Permutation & Combination questions


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Originally posted by EgmatQuantExpert on 04 Apr 2018, 05:38.
Last edited by EgmatQuantExpert on 13 Aug 2018, 06:45, edited 7 times in total.
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Re: A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals  [#permalink]

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New post 04 Apr 2018, 06:26
2
Signals can be generated by selecting 2 colours and rearranging them.
out of 4 colors: Red, Blue, Green, and Yellow, 2 can be selected in 4C2 ways = 6ways
arrangement = 2! = 2 ways.
No of signals = 6*2 = 12 ways.

Answer D
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Re: A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals  [#permalink]

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New post 04 Apr 2018, 19:33
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EgmatQuantExpert wrote:
Learn when to “Add” and “Multiply” in Permutation & Combination questions- Exercise Question #2

A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals are made by combining any \(2\) of these \(4\) colors. How many unique signals can the traffic signal show if the order of the colors defines unique cases?

A) 3
B) 6
C) 8
D) 12
E) 24

Permutation
Initially, we choose two colors from four.

The number of ways in which we can "choose" or "select" \(r\) objects from a group of \(n\) objects indicates a combination.

BUT order matters: "how many UNIQUE signals" and "the order of colour defines unique."
How the two colors are arranged matters. If order matters, permutation is needed.

The number of arrangements of \(r\) objects taken \(n\) at a time = permutation

R, G, B, Y = colors
RG is different from GR

Permutation, \(_{n}P_{r}\)

Formula: \(\frac{n!}{(n-r)!}\)

\(_{4}P_{2}=\frac{4!}{2!}=\frac{4*3*2*1}{2*1}=\)12 unique arrangements

Fundamental Counting Principle:

For the first color, there are 4 choices
No matter which color gets chosen first, after that there are 3 colors from which to select for the second choice

__4___ * __3___ = 12 arrangements

R, G, B, Y:

R chosen first: RG, RB, RY
G chosen first: GR, GB, GY
B chosen first: BR, BG, BY
Y chosen first: YR, YG, YB

Answer D
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Re: A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals  [#permalink]

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New post 08 Apr 2018, 20:30

Solution



Given:
    • A traffic light has signals of 4 colours: Red, blue, green and yellow.
    • The traffic light will show a signal by combining 2 of these 4 colours.

To find:
    • We need to find the number of ways in which 2 colours can be chosen out of 4 colours.

Approach and Working:

Since the picking of 2 different colours gives a different signal, the ways of picking of 2 colours are dependent on each other.
Hence,
    • Total ways to pick colours=Number of ways to pick 1 colours from 4 colours and number of ways to pick another colour
    • Total ways= 4*3= 12

Hence, the correct answer is option D.

Answer: D

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Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
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Re: A traffic signal has 4 colors: Red, Blue, Green, and Yellow. Signals &nbs [#permalink] 08 Apr 2018, 20:30
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