difference in results Q) In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
Solution Available:
It is simply: 103C3 - 3(52C3) = 110551.
Amazed. Now see the explanation part. Out of total 250 marks, you want to score 150 marks. So number of ways of scoring 150 is equivalent to not scoring the remaining 100 marks. So let's say we have 4 boxes a, b, c, d such that capacity of a, b, c is 50 each and that of d is 100. And all 4 boxes are filled to its capacity and we are to remove 100 balls out of it. Does it sound better?
i.e. we are trying to find the whole number solution of: a + b + c + d = 100 with the restriction that none of a, b, c can be more than 50. OK. Now without restriction, number of whole number solution of above equation is given by 103C3. We need to subtract the cases when any one of a, b, c is more than 50 (i.e. 51 + x say, where x is a whole number). Remember no more than one of a, b, c can be greater than 50. Also I have taken it as 51 + x because that's how I ensure that one of a, b, c is certainly more than 50. And which one of a, b, c can be determined in 3 ways. Hence the final upfront expression.
what i tried:
60% marks in aggregate means a total of 150. i have assumed a, b, c the marks in the first 3 papers and d as the marks in the fourth paper. so, a+b+c+d = 150 under the restriction that a(max)=b(max)=c(max)=50 & d(max)=100. then, i figured out the total no. of ways as
C(153,3) - 3*C(102,3) - C(52,3) = 48076
C(153,3) is the total no. of ways of distributing 150 amongst a,b,c,d without any restrictions.now, i am looking to subtract the cases where any of a,b,c is more than 50, or d is more than 100. Assuming that a=51, we can distribute 99 among the 4 groups in C(102,3) ways. i have multiplied C(102,3) with 3 to take care of similar cases for b and c. Lastly, assuming d=101, we can distribute 49 among 4 groups in C(52,3) ways. these cases are to be subtracted as well.
this is how i figured out the solution. i have understood above explanation.He approached the sum by thinking about the 100 marks which have not been scored. now, when i am applying similar logic and thinking about the 150 marks which have been scored, then why are the answers not matching up?
 A: The reason you did not obtain the correct answer is that two of the restrictions can be violated simultaneously.
The number of outcomes is the number of solutions of the equation in the nonnegative integers
$$a + b + c + d = 150$$
under the restrictions that
\begin{align*}
a & \leq 50\\
b & \leq 50\\
c & \leq 50\\
d & \leq 100
\end{align*}
You correctly found that if there were no restrictions, then the number of solutions would be 
$$\binom{150 + 3}{3} = \binom{153}{3}$$
You also correctly found that the restriction $a \leq 50$ eliminates 
$$\binom{150 - 51 + 3}{3} = \binom{102}{3}$$
as do the restrictions $b \leq 50$ and $c \leq 50$.  Your calculation that the restriction that $d \leq 100$ eliminates 
$$\binom{150 - 101 + 3}{3} = \binom{52}{3}$$
solutions is also correct.  
What you did not take into account is that two of the three restrictions $a \leq 50$, $b \leq 50$, and $c \leq 50$ can be violated simultaneously.  By subtracting the $$3\binom{102}{3}$$ cases in which one of these conditions was violated, you subtracted each of the three cases in which two of these conditions were simultaneously violated twice.  We only want to subtract them once.  Therefore, we must add them back.  Two particular conditions can be violated simultaneously (say $a > 50$ and $b > 50$) in 
$$\binom{150 - 2 \cdot 51 + 3}{3} = \binom{51}{3}$$
ways.  Since there are three such pairs, we must add 
$$\binom{3}{2}\binom{51}{3}$$
to your answer to obtain the
$$\binom{153}{3} - \binom{52}{3} - \binom{3}{1}\binom{102}{3} + \binom{3}{2}\binom{51}{3} = 110,551$$
possible ways in which a student could receive an aggregate mark of $60\%$.  
Note that this argument employs the Inclusion-Exclusion Principle.
