Number of wrong answers

combinatoricspermutations

In a certain test $a_i$ students gave wrong answers to at least i questions, where $i= 1,2,3……k$ No student gave more than k wrong answers. The total number of wrong answers is

I wasn't able to start solving this. I tried subtracting 2-1 for exactly 1 answer but I didn't understand what was happening. The solution provided uses the same logic from the below sum of Therefore, total number of wrong answers contributed by $2^{n−i} −2^{n−i−1}$ students who answered i questions wrong is $(2^{n−i}−2^{n−i−1})i$. I have no clue how this was derived.

There is a similar question on the site already but it is the other way around and doesn't resolve my doubt.
To find number of questions when number of wrong answers is given

Best Answer

  • The number who gave at least $k$ wrong answers is $a_k$. Since no student gave more than $k$ wrong answers, the number who gave exactly $k$ wrong answers is $a_k$.

  • The number who gave at least $k-1$ wrong answers is $a_{k-1}$. Since $a_k$ students gave more than $k-1$ wrong answers, the number who gave exactly $k-1$ wrong answers is $a_{k-1}-a_k$.

  • The number who gave at least $k-2$ wrong answers is $a_{k-2}$. Since $a_{k-1}$ students gave more than $k-2$ wrong answers, the number who gave exactly $k-2$ wrong answers is $a_{k-2}-a_{k-1}$.

  • $\cdots$

  • The number who gave at least $1$ wrong answer is $a_{1}$. Since $a_2$ students gave more than $1$ wrong answer, the number who gave exactly $1$ wrong answers is $a_1-a_2$.

So the total number of wrong answers is $$1(a_1-a_2)+2(a_2-a_3) +\cdots+(k-1)(a_{k-1}-a_k)+k(a_k)$$ $$=a_1 +a_2(2-1)+a_3(3-2)+\cdots+a_k(k-(k-1))$$ $$=a_1+a_2+a_3+...+ a_k$$ $$=\sum\limits_1^k a_n$$

Related Question