Multiple choice

A certain question has the following possible answers.

  1. All of the below
  2. None of the below
  3. All of the above
  4. One of the above
  5. None of the above
  6. None of the above

Which answer is correct?


42 thoughts on “Multiple choice

  1. Nan, The difference between “e” and “f” is that “e” is above “f”, which makes “f” invalid.

  2. This is a great puzzle. Let’s write out the implications of the various options:

    a --> b & c & d & e & f
    b --> !c & !d & !e & !f
    c --> a & b
    d --> a | b | c
    e --> !a & !b & !c & !d &
    f --> !a & !b & !c & !d & !e

    Since a --> b & c, but b --> !c, !a.
    Since c --> a, !c.
    Since d --> a | b | c and !a and !c, d --> b. But b --> !d. Thus, !d.

    We can simplify a bit:

    b --> !e & !f
    e --> !b
    f --> !b & !e

    Let’s write these out very explicitly:

    b --> !e
    b --> !f
    e --> !b
    f --> !b
    f --> !e
    e --> !f (contrapositive)

    We can make satisfying assignments:
    Suppose e. Then !b and !f. OK.
    Suppose f. Then !b and !e. OK.
    Suppose b. Then !e and !f. OK.

    So, how to decide? Note that the conditions exhibit a perfect symmetry between e and f. Thus, any chain of logic that could let us conclude e could equally well let us conclude f. But, these options are mutually exclusive.

    Therefore, neither e nor f, but b instead.

  3. So, the answer to Nan’s question is: there is absolutely no difference (in the above formal sense) between e and f, and that is crucial in concluding b!

  4. Evan, you missed a couple implications
    d -> a | b | c
    !d -> !a & !b & !c -> !b
    b -> d

    And also
    f -> (!a & !b & !c & !d) & !e -> e & !e

  5. I think d -> a | b | c is wrong. That means “at least one of the above”, but it says “one of the above”, which I have to take to mean “exactly one of the above”. In that case it should be d -> (a & !b & !c) | (!a & b & !c) | (!a & !b & c).

    If we further assume that there is only one answer, which the question “which answer is correct?” (not “which answers are correct?) seems to imply, then d cannot be true, as that would mean two correct answers. In fact, it immediately disallows all but b, e, and f. f implies e, so f must be false (f also implies !e, so it’s self-contradictory anyway). a and b cannot both be true, meaning a itself cannot be true, and hence c is also false. Thus, if b is true, it is the only one of a, b, and c, so d is also true, which we already ruled out. So the only possibility left is e.

  6. I spent a few minutes on this and didn’t write anything down. My intuition said (e). However, in retrospect, your comments are very interesting very enlightening. The approach I used was to set (a), (c) and (f) to false and just cycled through the remaining eight permutations.

    Very nice question. Thanks for sharing.

  7. Lvps1000vm, I guess your point is that those implications should really be rather than -->?

    Aaron, I think you’re right about the meaning of choice d. But the logic leading to !a and !c still hold, so d --> (a & !b & !c) | (!a & b & !c) | (!a & !b & c) can still be simplified to d --> b, I think.

  8. Ha, of course. What I meant was

    Lvps1000vm, I guess your point is that those implications should really be [two-directional arrow] rather than –>?

  9. The answer can not be ‘b’ since ‘d’ would also be true.

    The answer is ‘e’:

    ‘a’ conflicts with ‘b’ so the answer isn’t ‘a’.
    ‘b’ is false since there is an answer below.
    Since ‘a’ and ‘b’ are not the answer, ‘c’ and ‘d’ are false as well.
    ‘e’ works since none of ‘a’, ‘b’, ‘c’, or ‘d’ are the answer.
    ‘f’ fails due to the fact that ‘e’ is true.

    L. Euler

  10. No, my point is that b implies both d and not d, hence b is impossible. Also f implies e and not e, hence f is impossible and you get only e left.

    e and f aren’t symmetrical.

    (The first part of what I wrote in my earlier comment, what lead to b -> d, is totally false, as Aaron noted. But I can save it anyway: if a and c are false, then !d -> !b, and then b -> d. And a and c are false)

  11. I’m in agreement with Aaron regarding option E.

    If B were true it would make D true, resulting in a contradiction. E and F confused me momentarily but if F were true, it would make E true as a subset meaning F would be in contradiction. Thus, E remains as the correct answer without resulting in contradiction.

  12. Lvps1000vm,

    Your claim f -> (!a & !b & !c & !d) & !e -> e & !e means (!a & !b & !c & !d) imply e. But this need not necessarily be, unless the implications are bidirectional.

    Nonetheless, I do not understand

    (The first part of what I wrote in my earlier comment, what lead to b -> d, is totally false, as Aaron noted. But I can save it anyway: if a and c are false, then !d -> !b, and then b -> d. And a and c are false)

    If the implication is bidirectional, then I appreciate !d --> !b. Otherwise, I don't see how you get from d --> b to !d --> !b.

  13. By a process of elimination:

    If (c) then (a) hence (b) hence not (c) which is a contradiction, therefore not (c).
    If (a) then (c) which contradicts the previous step, therefore not (a).
    If (d) then (b) (as neither (a) nor (c) – doesn’t matter whether it means ‘exactly one’ or ‘at most one’) hence not (d) which is a contradiction, therefore not (d).
    If (b) then (d) too which contradicts the previous step, therefore not (b).
    Thus (e) (and thus not (f)).

  14. To add to the above discussion, the implications are indeed bidirectional. For instance, if you picked answer (b) then that would imply answer (d) too.

    If you read the implications as one-directional then there are three equally correct answers: ‘just (b)’, ‘just (e)’ and ‘just (f)’.

  15. I’m going with e, because the following is a consistent assignment with e being the only true statement.

    a. All of the below:
    false because b, c, d, and f are false
    b. None of the below:
    false because e is true
    c. All of the above
    false because a and b are false
    d. One of the above
    false because a, b, and c are all false
    e. None of the above
    true because a, b, c, and d are all false
    f. None of the above
    false because e is true

  16. Decided to run this through SymPy’s SAT solver.

    In [1]: var(‘a b c d e f’)
    Out[1]: (a, b, c, d, e, f)

    In [2]: facts = [
    Equivalent(a, (b & c & d & e & f)),
    Equivalent(b, (~c & ~d & ~e & ~f)),
    Equivalent(c, a & b),
    Equivalent(d, ((a & ~b & ~c) | (~a & b & ~c) | (~a & ~b & c))),
    Equivalent(e, (~a & ~b & ~c & ~d)),
    Equivalent(f, (~a & ~b & ~c & ~d & ~e)),

    In [3]: list(satisfiable(And(*facts), all_models=True))
    Out[3]: [{e: True, c: False, b: False, a: False, f: False, d: False}]

    So it seems e is the only answer, assuming I got the facts correct. And it is important to use Equivalent (a bidirectional implication) rather than just Implies. If you only use -> (which I guess would mean that an answer not being chosen doesn’t necessarily mean that it isn’t true), then ‘none’, b, and f are also “solutions”.

    Also, if I replace the d fact with Equivalent(d, a | b | c), the result is the same. So it seems that the interpretation of “one” both in terms of choice d and in terms of how many choices there are is irrelevant.

    Thanks for the fun problem. I hope others took the time to solve this in their head before reading the comments.

  17. d -> a | b | c is certainly not wrong. It’s less information than you could represent, but it’s perfectly true. “One of the above” always implies “one or more of the above”.

    It really bothers me to see people saying b is possible.

    a -> b & c
    b -> !c
    therefore, a -> c & !c
    therefore, !a

    c -> a
    therefore, !c

    b -> !d
    !a & b & !c -> d
    !a & !c
    therefore, b -> !a & b & !c
    therefore, b -> d & !d
    therefore, !b.

  18. Here’s the easy solution :

    • Suppose a True. Then b is True which implies a is False which is absurd. Therefore a is False.
    • Suppose c True. Then a is True which is wrong. Therefore c is False.
    • Suppose d True. Then either a,b,c must be True but a=c=False thu b is True which implies d is False which is absurd. Therefore d is False.
    • Suppose b is True. Then d is False thus a=b=c=False which is absurd. Therefore b is False.
    • a=b=c=d=False therefore e is True.
    • e is True therefore f is False.

  19. Oops I meant e naturally. So here’s the condensed and corrected version :

    a => e => !a . Absurd. Thus a is wrong.
    c => a but a is wrong. Thus c is wrong.
    d => (a or b or c) but a,c are wrong, thus d=>b=>!d. Absurd. Thus d is wrong, which means b is also wrong.
    a,b,c,d are wrong thus e is true.
    e is true thus f is wrong.

  20. I feel like some of the commenters are reading “none of the above” as “none on this list” whereas I read it as none of the options above said option which to me makes the answer more obvious.

  21. This was obviously e, not sure what the big deal is. Simple process of elimination as Jio15 said, easily done mentally.

  22. I agree with (almost) everyone else that the answer is (e) and it’s not terribly hard. But there’s something a bit iffy about this sort of reasoning.

    What’s the answer to this multiple-choice problem? “(a) Not a.” Clearly no choice you can make leads to a consistent result.

    In the absence of any actual assurance that there is a consistent answer you’re not really entitled to eliminate answers that lead to inconsistency. See, e.g., this blog post by Mark-Jason Dominus for more on this theme.

  23. Nice post. My answer is “e”.
    If you assume “e” is true, then all of the above options must be false, and the last can not be true since “e” is true.

  24. g, I think this puzzle is different. In that puzzle, there are two independent things, the labels, and the treasure. The box that contains the treasure need not be the one with the true label. If you were to write out logical clauses for it, you would need at least three variables, two for the labels, and one for the treasure. In particular, in that puzzle, you know that only one box contains a treasure, but you don’t know that only one box has a true label. They could both be true, or neither (which is the trick of the argument). It’s like the two guards at the gate problem except if you only mention that one guard always lies and don’t explicitly state that the other always tells the truth. They could both be lying.

    In this puzzle, the selected choice and the true choice are necessarily one and the same.

  25. Answer: e
    if ‘e’ is incorrect, neither ‘f’ nor ´a’ or ‘c’ can be correct; additionally,
    since there is only one correct answer, ‘b’ is contradictory to ‘d’ (either both
    would be true or false).

  26. It’s not possible to give an answer to this.

    The puzzle specifically says that these are the answers to “a certain question”, which we are never told. The last sentence is clearly the question of this puzzle, i.e. “Which answer is correct [for the original, unknown question]?”, and not the actual original question.

    The question could have been e.g. “Which of the following sentences has the most L’s in it?”, in which case the correct answer is a, or “Which answer starts with ‘b’?”, in which case the answer is d. Since we don’t know the question, we can’t know the answer.

  27. Just had to comment since both Jeff Scott Brown, which got me hooked on Groovy and Grails through this video , and an impersinator for my all time favorite human being Leonard Euler ( commented! :-)

    I actually solved this rigorously with boolean algebra brute force!
    The paper and pen equivalence to Aaron Meurer’s SymPy’s SAT solver.
    Checking all 64 (2^6) combinations…
    a b c d e f – holds for (output)
    0 0 0 0 0 0 – b,e,f
    0 0 0 0 0 1 – e,f
    0 0 0 0 1 0 – e
    0 0 0 0 1 1 – e
    60 more rows…

    Since there is a claim that only one solution exist, there should be one sole match between the table input and the output. The third row, e, is the only match and the answer!

  28. the last statement is wrong.
    regardless of the question, if you state that a is the answer, it also states that the statement of a is correct. but we can show a will not be correct regardless of the question, so we can drop the philosophy. its a simple mathematical problem in logic, and can be solved by math tools such as Modus Ponens and/or Modus Tollens.

    similar to what a couple of you wrote here, you can assume an answer and show it contradicts itself.
    example: f -> !e -> at least one of (a,b,c,d) -> !f
    therefore f -> !f so f contradicts itself. we conclude: => !f
    a -> f -> !a => !a
    b – a bit more complex, so if you have a simple way – , I’ll be happy to
    find your bug :), but it can be simple if we show !c and !d first.
    …skipping to c:
    c -> a -> !a (as shown 3 lines above) -> !c => !c
    by definition d -> one of (a,b,c) but we shown !a and !c so :
    d -> one of (b) => d-> b.
    /* that is called BTW Modus Tollens: (d->c & !c ) -> !d similar for !a */
    we know b -> !d, so d->b->!d => !d
    so now we have proven: !a, !c, !d, !f moving on to b.
    lets assume b is correct. so (b & !a & !c) => one of (a,b,c) ->d.
    in short: b->d. but we know !d so using Modus Tollens:
    (b->d & !d ) -> !b

    so now we had shown !a, !b, !c, !d, !f . Lets check e:
    as we know !a, !b, !c, !d => e (by definition of e)
    so e is the correct answer.

    thanks for a lovely question.

  29. @Tor: John’s post provides a specific example showing how poorly-constructed multiple-choice answers may contain inconsistencies that permit some answers to rule out other answers.

    When helping students with multiple-choice questions, the first step after reading the question and the answers is to eliminate all answers that cannot possibly be correct (or, conversely, must be incorrect). Then repeat until the list of answers fails to shrink.

    Surprisingly often, the correct answer may be deduced without needing to fully work the problem.

    This post contains an extreme case, where the problem statement itself is not even necessary!

  30. import Data.SBV
    choice = allSat $ do
                 [a, b, c, d, e, f] <- sBools ["a", "b", "c", "d", "e", "f"]
                 solve [ a .== bAnd [b, c, d, e]               -- a. All of the below
                       , b .== bAnd (map bnot [c, d, e])       -- b. None of the below
                       , c .== bAnd [a, b]                     -- c. All of the above
                       , d .== foldr1 () [a, b, c]          -- d. One of the above
                       , e .== bAnd (map bnot [a, b, c, d])    -- e. None of the above
                       , f .== bAnd (map bnot [a, b, c, d, e]) -- f. None of the above

    We have:

    *Main> choice
    Solution #1:
      a = False
      b = False
      c = False
      d = False
      e = True
      f = False
    This is the only solution.

    It appears e is the way to go.

  31. I reformatted your previous comment. The trick is to wrap everything in pre tags, not code tags.

  32. It is not written anywhere which order we should read the answer
    of question I have one argument here who is deciding that first we should read the option E while if I read F firs t then I find E is false now since I have to choose only one option I will go with F because I read first F rather than E

  33. Since E and F both statement are identical we can’t rule out any option
    It is statement rather than any figure Both Statement carry equall weight so ignoring any one out of E & F is not logic
    Why can’t we assume option F some one read first and make his decision F rather than going for mathematical calculation

  34. its very simple when we change in numeric question example
    what is 2+2

    select one answer
    Can we rule out any option between E and F ?

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