In a paper, there are four multiple-choice questions. Each has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

examrobotsa's picture
Q: 112 (IAS/2006)
In a paper, there are four multiple-choice questions. Each has five choices with only one choice for its correct answer. What is the total number of ways in which a candidate will not get all the four answers correct?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,4,6,1,3,4,2

keywords: 

{'choice questions': [0, 0, 1, 0], 'choices': [0, 0, 1, 0], 'total number': [0, 0, 3, 0], 'choice': [0, 1, 2, 0], 'candidate': [1, 0, 0, 0]}

Each question on the paper has one correct answer out of five options. That means, for each question there are four ways to answer it incorrectly.

Option 1: 19 is not relevant in this context because it doesn`t take into account the multiple options for each question and the possibility for incorrectly answering more than one question.

Option 2: 120 usually represents permutations such as 5 factorial (5*4*3*2*1), but here it doesn`t apply considering each question allows either right or wrong with more weightage for wrong answers.

Option 3: 624 represents the total number of ways a candidate can not get all four answers correct. The breakdown is: for each question there are 4 wrong answers and 1 right (total 5 options). So, 5 options for each of the 4 questions is 5^4 = 625. But, this count includes the one way to get all answers correct. So, we subtract that one way from the total 625, hence 624 ways of not getting all right.

Option 4: 1024 is the result of 4^4 = 256, which is the total number of possibilities for 5 choices for 4 questions, without recognizing the constraint of

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