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Four pool balls—A, B, C, D—are randomly arranged in a straig

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Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink] New post 08 May 2019, 04:28
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Four pool balls—A, B, C, D—are randomly arranged in a straight line. What is the probability that the order will actually be A, B, C, D ?

(A) \(\frac{1}{4}\)

(B) \(\frac{1}{4C_4}\)

(C) \(\frac{1}{4P_4}\)

(D) \(\frac{1}{2!}\)

(E) \(\frac{1}{3!}\)
[Reveal] Spoiler: OA

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Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink] New post 12 May 2019, 20:14
Can someone please help me with this question?
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Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink] New post 13 May 2019, 02:18
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A ball cannot exist in two slots, so repetition is not allowed.


Each ball is given a different identity A, B, C, and D, so there are no indistinguishable objects.

Here, n = 4 (number of balls to arrange) in r = 4 (positions). We know the problem type, and the formula to use. Hence, by Formula 2, the number of arrangements possible is \(4P_4\) , and {A, B, C, D} is just one of the arrangements. Hence, the probability is 1 in \(4P_4\) , or \(\frac{1}{4P_4}\).

The answer is (C).
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Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink] New post 13 May 2019, 07:09
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An easy mnemonic way to distinguish between Permutations and Combinations is that a Permutation is a permanent unique iteration where order matters, such as this situation where A | B | C | D is not the same outcome as C |B | D | A. Whereas, a Combination is an iteration where order does not matter and can be changed for instance if we were to select four candidates for two identical job openings this would be a combination since it would not matter whether a person were selected for opening one or two.

As it pertains to this specific question, we should be able to conceptually determine that there is only one method for selecting exactly A | B | C | D as requested. Then, we can determine that there would be four ways to select the first ball, three for the second ball, two for the third ball, and only one for the fourth ball. Since we are selecting each of those four balls, it is necessary to multiply each of the possible outcomes together to determine the total number of possibilities. Therefore, there would be 4 x 3 x 2 x 1 or 4! = 16 total possibilities and we are exactly seeking one of those choices, so you may be able to determine that the correct answer should be 1/16. However, of course neither 1/16 nor 1/4! are available options, so we just recall our mnemonic to determine that this scenario is a Permutation that would be permanently set without the flexibility to change spots to inform that we would need to select the P formula in choice C rather than the C formula in choice B.
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Re: Four pool balls—A, B, C, D—are randomly arranged in a straig [#permalink] New post 26 Nov 2019, 01:53
Carcass wrote:
A ball cannot exist in two slots, so repetition is not allowed.


Each ball is given a different identity A, B, C, and D, so there are no indistinguishable objects.

Here, n = 4 (number of balls to arrange) in r = 4 (positions). We know the problem type, and the formula to use. Hence, by Formula 2, the number of arrangements possible is \(4P_4\) , and {A, B, C, D} is just one of the arrangements. Hence, the probability is 1 in \(4P_4\) , or \(\frac{1}{4P_4}\).

The answer is (C).


What do you mean by formula 2?
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Re: Four pool balls—A, B, C, D—are randomly arranged in a straig   [#permalink] 26 Nov 2019, 01:53
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