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If Jeff has four movies, and must choose to watch either 1, 2, or 3 di

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Bunuel
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If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink] New post   02 Jul 2020, 02:07
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43% (01:17) correct 57% (01:27) wrong based on 61 sessions
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If Jeff has four movies, and must choose to watch either 1, 2, or 3 different movies, which of the following represents a possible number of different arrangements of movies that Jeff could watch?

I. 4
II. 12
II. 24

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III
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E
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nick1816
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Re: If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink] New post   02 Jul 2020, 05:08
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If Jeff chooses 1 movie, total possible arrangements = 4

If Jeff chooses 2 movies, total possible arrangements = 4*3 = 12

If Jeff chooses 3 movies, total possible arrangements = 4*3*2 = 24

E



Bunuel wrote:
If Jeff has four movies, and must choose to watch either 1, 2, or 3 different movies, which of the following represents a possible number of different arrangements of movies that Jeff could watch?

I. 4
II. 12
II. 24

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III
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Neha2404
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Re: If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink] New post   02 Jul 2020, 06:00
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IMO-E
Jeff has 4 movies options

1jeff watch only one movie =. 4C1 =4
2- jeff watch 2 movies =. 4C2. *2! = 12
3-jeff watch 3 movies =. 4C3 *3! =24

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Re: If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink] New post   02 Jul 2020, 06:18
Let the movies be ABCD
For 1 movie he can choose either A or B or C or D so 4 is right.

For 2 movies he can choose 4 of the above , and 3 for the second movie for 4*3 , so 12 is good.

For 3 movies same logic 4 ( a or b or C or D)* 3 ( minus the 1st choice) *2 ( minus 1st+2nd choices) for a total of 24.

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Re: If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink] New post   02 Jul 2020, 06:19
Let the movies be ABCD
For 1 movie he can choose either A or B or C or D so 4 is right.

For 2 movies he can choose 4 of the above , and 3 for the second movie for 4*3 , so 12 is good.

For 3 movies same logic 4 ( a or b or C or D)* 3 ( minus the 1st choice) *2 ( minus 1st+2nd choices) for a total of 24.

Posted from my mobile device is

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Re: If Jeff has four movies, and must choose to watch either 1, 2, or 3 di [#permalink]
02 Jul 2020, 06:19
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