Combinatoric question about difficult and easy problems

Question

Three children attempted one hundred problems. Each child solved sixty problems, and every problem was solved by at least one child. A problem is called difficult if it was only solved by one child, and called easy if it was solved by all three children. How many more easy problems were there than difficult problems?

I have tried to make progress on this question but I always end up with more difficult questions than easy questions. I got this question from a homework sheet from my teacher but I don't know the textbook it is from. I have been working on this question for over an hour and I have made no progress.

I used the example that:

Child 1 solved questions $1-60$, but not $61-100.$

Child 2 solved $41-100$ but not $1-40$ (so each question is at least solved by one child)

Child 3 solved questions $1-60$ but not $61-100.$

So the questions where all three children solved them was $41-60,$ but questions $61-100$ was where only one person solved it. This means that there is $40$ hard questions and $20$ easy questions, meaning there was $-20$ more easy problems than hard questions.

It is quite an interesting problem because it seems easy to explain, but not easy to solve. I did get an answer of $-20 $ when I tried an example, but it didn't seem right to me because I did not expect the answer to be negative, so I thought my method was wrong. I want to know how to solve these types of questions in the future.

Answer 1

Let $d$ be the number of difficult problems, $e$ the number of easy problems and $m$ the medium problems.

There are $100$ problems, so $d+m+e = 100$.

Each child solves $60$ questions, so from the definition of easy, medium, difficult we have $d+2m+3e=3\cdot60=180$.

There are 3 variables and only two equations, so this is not enough to determine each variable precisely. The question however asks only about $d$ and $e$, so let's see what happens when we eliminate the variable $m$. Multiplying the first equation by 2 and subtracting the second equation gives:

$$2(d+m+e)-(d+2m+3e) = 2\cdot100-180$$ which simplifies to $$d-e = 20$$

There are therefore $20$ fewer easy questions than difficult questions. The original problem statement seems to assume that there are more easy questions than difficult ones, so it is unclear if there is a mistake in the question or whether they expect the answer $-20$.

Answer 2

Let $A$ be the set of questions solved by the first child, $B$ the set solved by the second child and $C$ for the third.

You are told the following information: "Three children attempted 100 problems... and every problem was solved by at least one child" $|A\cup B\cup C|=100$

"Each child solved 60 problems" $|A|=|B|=|C|=60$

We know from inclusion-exclusion that $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$

Plugging in numbers into inclusion-exclusion we have $100 = 180 - |A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$ and so $|A\cap B|+|A\cap C|+|B\cap C| - |A\cap B\cap C| = 80$

What we want to find is the difference of the amount of problems solved by all students versus solved by only a single student. The amount solved by all is $|A\cap B\cap C|$. The amount solved only by $A$ would be $|A|-|A\cap B|-|A\cap C|+|A\cap B\cap C|$, seen similar to inclusion-exclusion. Similarly for $B$ and $C$.

So, the expression you are after in the end is:

$$|A\cap B\cap C|-(A-|A\cap B|-|A\cap C|+|A\cap B\cap C|) - (B-|A\cap B|-|B\cap C|+|A\cap B\cap C|)-(C-|A\cap C|-|B\cap C|+|A\cap B\cap C|)$$

$$=-2|A\cap B\cap C| - |A|-|B|-|C|+2|A\cap B|+2|A\cap C|+2|B\cap C|$$

$$=2(|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|) - |A|-|B|-|C|$$

Plugging in numbers, we have then

$$=2(80)-60-60-60 = 160-180=-20$$

That is, there were $20$ more hard questions (solved by only a single student) than there were easy questions (solved by all students)

Note... we don't care (nor can we determine exactly) how many hard questions there were or how many easy questions there were... but despite this we were able to find the difference in size between them.

For instance, perhaps the students all solved the first $40$ problems, but then each solved a different twenty of the remaining, giving a total of $40$ easy questions at $60$ hard questions. Alternatively, perhaps student $A$ solved problems $1$-$60$, student $B$ solved problems $1$-$20$ and $61$-$100$, and student $C$ solved problems $41$-$100$ for a total of zero easy questions and $20$ hard questions.

Answer 3

Letting $A$, $B$, and $C$ be the sets of questions answered correctly by the each of the three students, the Generalized Inclusion-Exclusion Principle says the following:

The number of questions answered by exactly $0$ students is $$ \begin{align} \overset{\substack{\text{answered by none}\\\downarrow}}{\color{#C00}{0}}&=\binom{0}{0}\overbrace{|A\cup B\cup C|}^{N(0)=100}-\binom{1}{0}\overbrace{(|A|+|B|+|C|)}^{N(1)=180}\\ &+\binom{2}{0}\overbrace{(|A\cap B|+|B\cap C|+|C\cap A|)}^{N(2)}-\binom{3}{0}\overbrace{|A\cap B\cap C|}^{N(3)=\text{easy}} \end{align} $$ The number of questions answered by exactly $1$ student is $$ \begin{align} \color{#090}{\text{difficult}} &=\binom{1}{1}\overbrace{(|A|+|B|+|C|)}^{N(1)=180}-\binom{2}{1}\overbrace{(|A\cap B|+|B\cap C|+|C\cap A|)}^{N(2)}\\ &+\binom{3}{1}\overbrace{|A\cap B\cap C|}^{N(3)=\text{easy}} \end{align} $$ Therefore, cancelling $N(2)$ in both, $$ \begin{align} \color{#090}{\text{difficult}}+2\cdot\color{#C00}{0} &=\color{#090}{(180-2N(2)+3\,\text{easy})}+2\,\color{#C00}{(100-180+N(2)-\text{easy})}\\ &=\text{easy}+20 \end{align} $$ Thus, there are $20$ more difficult questions than easy questions.