In a Question paper there are 4 multiple choice questions

Very stupid indeed on my part...need to sleep i guess......

Bunuel,

I could not understand how did you calculate that there will be only one case when the candidate answered all the four questions correct.

There is only one correct answer to each question. So, there is 1*1*1*1=1 way to answer all 4 questions correctly.

IMO:

4 questions with each having 5 options.
There are total 5 ^4 patterns in which a student can respond.

Hence total ways of responding-5^4= 625
Now, there is 1 way which contains all 4 answers incorrect by a student.

Hence we need to subtract that-
Hence Answer = 5^4 -1= 625-1 = 624

IMO option D is answer

We can use the following equation:

Total number of ways to answer all questions - number of ways to get all questions correct = the total number of ways in which a candidate will not get all four answers correct

5 x 5 x 5 x 5 - 1 x 1 x 1 x 1 = 625 - 1 = 624

Answer: D

Let's first determine the TOTAL number of ways the test can be completed.
The 1st question can be answered in 5 different ways (A, B, C, D, or E).
The 2nd question can be answered in 5 different ways (A, B, C, D, or E).
The 3rd question can be answered in 5 different ways (A, B, C, D, or E).
The 4th question can be answered in 5 different ways (A, B, C, D, or E).

By the Fundamental Counting Principle (FCP), the total number of ways we can complete test = (5)(5)(5)(5) = 625 ways

So, there 625 possible outcomes
Among those 625 possible outcomes, ONLY 1 outcome is such that all four questions ARE answered correctly.
This means, in the remaining 624 outcomes, the four questions are NOT all answered correctly.

Answer: D

Cheers,
Brent

Is there another, albeit less elegant, way to solve this problem?

You cannot answer all 4 questions correctly, so I was trying to add the combinations of 3 correct, 2 correct, 1 correct and none correct and adding them together.

For example, for 3 correct, you have 3*(5C1) (3 correct answers) * 1*(5C4) (for the incorrect answer).

Can't seem to get the correct answer using this method though. Thoughts?

I got this question wrong because I attempted to use factorials (i.e., 5! * 4!).

Can someone please explain when it is appropriate to use factorials, and when we should simply multiply the number of options together repeatedly (like in this question)?

Thanks!

Hi Bunuel,

Why shouldn't we include the internal arrangement of those 4 questions? I see that in your method, the order of the questions doesn't matter. How do we know when to consider the order and when not to?

Thanks in advance.

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