# Math Help - 4 answers to 4 questions

1. ## 4 answers to 4 questions

3. In an examination a candidate is given the four answers to four questions but is not told which answer applies to which question. He is asked to write down each of the four answers next to its appropriate question.

b). show that there are 6 ways of getting just two answers in the correct places.

c). If a candidate guesses at random where the four answers are to go and X is the number of correct guesses he makes, draw up the probability distribution for X in tabular form.

this question has been asked before in an old post, with no answer.
Balls in bags, 4 answers to 4 exam questions.

For b), is it $^4C_2$? it is just a matter of choosing which 2 of the 4 are correct. for the remaining 2, there is only one way that they can be wrong, that is when the 2 are swapped. so,$^4C_2$. but i myself am not very persuaded by this approach. wondering if there is a better way which can be applied to other similar cases (because i don't think i can use this approach to do other similar cases, like finding the number of ways of getting 5 answers correct if there are 11 questions and 11 answers. btw it would be cool if somebody can tell me how to do this too)

For c) i spend so much time trying this, and i just cannot find the number of ways for 0 correct answer and 1 correct answer! this is not just the simple arranging objects that i have been doing, this involves arranging objects into fixed spaces. fyi the answer for 0 correct spaces is 9 ways and 1 correct answer is 8 ways. i googled and found this reply(for 0 correct answers):

"This
is just the number of permutations of the set {1,2,3,4} that change eachelement. Each permutation that moves all elements can be represented in the form (1 x y z) or (1 x)(y z), where x = (1), z = (y), etc.

So the # of possible permutations that change all elements is just the number of possible configurations of (x y z) = plus the number of configurations of (1 x)(y z) which is 3, for a total of 9 ways."

but i do not understand the notation he is using. somebody please explain this

2. ## Re: 4 answers to 4 questions

Originally Posted by muddywaters
3. In an examination a candidate is given the four answers to four questions but is not told which answer applies to which question. He is asked to write down each of the four answers next to its appropriate question.

b). show that there are 6 ways of getting just two answers in the correct places.
This is pretty straight forward. Writing "T" for a correct answer and "F" for an incorrect answer this is just asking how many way you can arrange "TTFF" and that is $\frac{4!}{2!2!}= \frac{24}{4}= 6$.

c). If a candidate guesses at random where the four answers are to go and X is the number of correct guesses he makes, draw up the probability distribution for X in tabular form.

this question has been asked before in an old post, with no answer.
Balls in bags, 4 answers to 4 exam questions.

For b), is it $^4C_2$? it is just a matter of choosing which 2 of the 4 are correct. for the remaining 2, there is only one way that they can be wrong, that is when the 2 are swapped. so,$^4C_2$. but i myself am not very persuaded by this approach. wondering if there is a better way which can be applied to other similar cases (because i don't think i can use this approach to do other similar cases, like finding the number of ways of getting 5 answers correct if there are 11 questions and 11 answers. btw it would be cool if somebody can tell me how to do this too)

For c) i spend so much time trying this, and i just cannot find the number of ways for 0 correct answer and 1 correct answer! this is not just the simple arranging objects that i have been doing, this involves arranging objects into fixed spaces. fyi the answer for 0 correct spaces is 9 ways and 1 correct answer is 8 ways. i googled and found this reply(for 0 correct answers):

"This
is just the number of permutations of the set {1,2,3,4} that change eachelement. Each permutation that moves all elements can be represented in the form (1 x y z) or (1 x)(y z), where x = (1), z = (y), etc.

So the # of possible permutations that change all elements is just the number of possible configurations of (x y z) = plus the number of configurations of (1 x)(y z) which is 3, for a total of 9 ways."

but i do not understand the notation he is using. somebody please explain this
Similarly, "0 correct answers" is "FFFF" and there is only 1 way to do that. "1 correct answer " is "TFFF" and there are $\frac{4!}{1!3!}=\frac{24}{6}= 4$ ways to arrange that (TFFF, FTFF, FFTF, FFFT). "2 correct answers" is 6 as above. "3 correct answers" if just "FTTT" which again gives 4, and "4 correct answers", "TTTT" can be done in just one way.

3. ## Re: 4 answers to 4 questions

Originally Posted by muddywaters
3. In an examination a candidate is given the four answers to four questions but is not told which answer applies to which question. He is asked to write down each of the four answers next to its appropriate question.

For c) i spend so much time trying this, and i just cannot find the number of ways for 0 correct answer and 1 correct answer! this is not just the simple arranging objects that i have been doing, this involves arranging objects into fixed spaces. fyi the answer for 0 correct spaces is 9 ways and 1 correct answer is 8 ways. i googled and found this reply(for 0 correct answers):

"This is just the number of permutations of the set {1,2,3,4} that change eachelement. Each permutation that moves all elements can be represented in the form (1 x y z) or (1 x)(y z), where x = (1), z = (y), etc.

So the # of possible permutations that change all elements is just the number of possible configurations of (x y z) = plus the number of configurations of (1 x)(y z) which is 3, for a total of 9 ways."
I take it to mean that there are four distinct questions and four distinct answers.
I see this as a matter of counting derangements.

There are $4!=24$ ways to answer the four questions.
There are $9$ derangements of four items. Therefore there are nine ways to get none right.

Because there are $2$ derangements of three items and four ways to pick one. Therefore there are eight ways to get exactly one right.
$9+8+6+1=24$

4. ## Re: 4 answers to 4 questions

Originally Posted by HallsofIvy
This is pretty straight forward. Writing "T" for a correct answer and "F" for an incorrect answer this is just asking how many way you can arrange "TTFF" and that is $\frac{4!}{2!2!}= \frac{24}{4}= 6$.

Similarly, "0 correct answers" is "FFFF" and there is only 1 way to do that. "1 correct answer " is "TFFF" and there are $\frac{4!}{1!3!}=\frac{24}{6}= 4$ ways to arrange that (TFFF, FTFF, FFTF, FFFT). "2 correct answers" is 6 as above. "3 correct answers" if just "FTTT" which again gives 4, and "4 correct answers", "TTTT" can be done in just one way.
I think u missed out the point that the student is given the four answers to the four questions but is not told which answer is for which question. There is not only 1 way to get all the answers wrong. For example the questions are $A, B, C, D$ and the answers are $a, b, c, d$ respectively. Entering$c, d, a, b$ and $d, c, b, a$ are just 2 of the several ways to put in 4 wrong answers.
What i have so far is that there are $4!=24$ways to answer in total. There are $0$ ways to get only 3 answers correct, because if so the fourth will be correct. There are $^4C_2=6$ ways to get 2 answers correct (as mentioned above, i am wondering if theres a better or more logical way to find a better way to derive $6$ than $^4C_2$).

The problem is that i dont know how to calculate the number of ways to answer 0 corectly and answer 1 correctly, however the answer is 9 and 8 respectively.

5. ## Re: 4 answers to 4 questions

Originally Posted by muddywaters
I think u missed out the point that the student is given the four answers to the four questions but is not told which answer is for which question. There is not only 1 way to get all the answers wrong. For example the questions are $A, B, C, D$ and the answers are $a, b, c, d$ respectively. Entering$c, d, a, b$ and $d, c, b, a$ are just 2 of the several ways to put in 4 wrong answers

6. ## Re: 4 answers to 4 questions

Originally Posted by Plato
I take it to mean that there are four distinct questions and four distinct answers.
I see this as a matter of counting derangements.

There are $4!=24$ ways to answer the four questions.
There are $9$ derangements of four items. Therefore there are nine ways to get none right.

Because there are $2$ derangements of three items and four ways to pick one. Therefore there are eight ways to get exactly one right.
$9+8+6+1=24$
Yes, i didnt see this post before i replied.

How do i know that there are 9 derangements?? I have not learned that. This just popped up among questions involving simple permutation and combination. I dont believe im expected to use the formula in the wiki page u linked, if thats even the formula haha, didnt quite absorb all of that

7. ## Re: 4 answers to 4 questions

Originally Posted by muddywaters
How do i know that there are 9 derangements?? I have not learned that. This just popped up among questions involving simple permutation and combination. I dont believe im expected to use the formula in the wiki page u linked, if thats even the formula haha, didnt quite absorb all of that
Go to this webpage.

8. ## Re: 4 answers to 4 questions

for future reference and for anybody who's new to this like me, i just watched this and it is helpful