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28 Jan 2021, 15:29
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D
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Difficulty:
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84% (01:18) correct
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There are 6 multiple choice questions in an examination. How many sequences of answers are possible if the first 3 questions have four choices each and the remaining 3 questions have three choices each?
A. 21
B. 36
C. 81
D. 1728
E. 46656
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A. 21
B. 36
C. 81
D. 1728
E. 46656
The Ans to this is 4*4*4*3*3*3 = 1728 possible answers.
MY QUERY IS - do we take it as 4C1*4C1*4C1*3C1*3C1*3C1
Or is it 4P1*4P1*4P1*3P1*3P1*3P1 ??
Basically is it a combination problem or a permutation problem?
Both give exactly same abswers but what is correct conceptually?
MY QUERY IS - do we take it as 4C1*4C1*4C1*3C1*3C1*3C1
Or is it 4P1*4P1*4P1*3P1*3P1*3P1 ??
Basically is it a combination problem or a permutation problem?
Both give exactly same abswers but what is correct conceptually?
29 Jan 2021, 08:27
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There are 6 multiple choice questions in an examination. How many sequences of answers are possible if the first 3 questions have four choices each and the remaining 3 questions have three choices each?
Choices for the first question = 4
Choices for the first and second question = 4 * 4 = 16
Similarly, Choices for the first to sixth question = 4 * 4 * 4 * 3 * 3 * 3= 1728
Choice D is the answer.
Choices for the first question = 4
Choices for the first and second question = 4 * 4 = 16
Similarly, Choices for the first to sixth question = 4 * 4 * 4 * 3 * 3 * 3= 1728
Choice D is the answer.
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05 Feb 2021, 08:20
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Possible Sequences for First 3 questions = \(4^3\)
Possible Sequences for Last 3 questions = \(3^3\)
Total Sequences = \(4^3\)*\(3^3\) = \(12^3\) = 1728
Option D
Possible Sequences for Last 3 questions = \(3^3\)
Total Sequences = \(4^3\)*\(3^3\) = \(12^3\) = 1728
Option D
