A Twisted Problem From The IOQM 2025

image containing math problem statement
Here is a slightly twisted one from the IOQM (Indian Olympiad Qualifier In Mathematics) 2025. Make sure you know your Permutation & Combination well.
Problem contributed by debosmita1729
-----------book page break-----------
The value of $d$ ranges from $4$ to $98$.

When we choose $d = 4$, $e$ can be chosen in $\displaystyle \binom {95}{1}$ ways and $a, b, c$ can be chosen in $\displaystyle \binom {3}{3}$ ways.
 
Therefore, sum of $d$ when $d = 4$ is $4 \times \displaystyle\binom{95}{1} \times \binom{3}{3}$

Similarly,
sum of $d$ when $d = 5$ is $\displaystyle5 \times \binom{94}{1} \times \binom{4}{3}$
 
sum of $d$ when $d = 6$ is $\displaystyle6 \times \binom{93}{1} \times \binom{5}{3}$
$\vdots$
sum of $d$ when $d = 98$ is $\displaystyle98 \times \binom{1}{1} \times \binom{97}{3}$
 
Total Sum = $\displaystyle 4 \times \binom{95}{1} \times \binom{3}{3} + 5 \times \binom{94}{1} \times \binom{4}{3} + ... + 98 \times \binom{1}{1} \times \binom{97}{3}$

-----------book page break-----------
$ = 4 \times \displaystyle \sum_{r = 1}^{95} \dfrac{99 - r}{4} \times r \times \binom{98 - r}{3}$
 
$= \displaystyle 4 \times \sum_{r = 1}^{95} \binom{99 - r}{4} \times r$
 
$= 4 \times \displaystyle \binom{100}{6}$

Total Number of Tuples  $ = \displaystyle \binom{99}{5}$
 
$D = \displaystyle \dfrac{4 \times \binom{100}{6}}{\binom{99}{5}}$
 
$\Rightarrow D = \dfrac{200}{6}$
 
$\therefore\displaystyle \left\lfloor{\dfrac{200}{6}}\right\rfloor = 66$