A Marta Sved combinatorics problem: scheduling two-day fencing and javelin competitions, and a one-day celebration, in a month with 31 days 
A university wants to organize a fencing and javelin throwing competitions during the month of May that counts 31 days.
Each of these competitions will take 2 days, not necessarily consecutive, and since the University knows that none of the competitors is involved in both pursuits, fencing and javelin throwing could be done on different days or on the same day.
BUT after the competitions, a feast (or a celebration day if you prefer) must take place in the university before the month ends.
In how many ways could you draw up a timetable for the events?

Let's call C-day the celebration day, F-days and J-days the fencing and javelin days. One possible solution, knowing that the C-day cannot be held before  the 3rd of the month, would be given by the following sum:
\begin{equation}
\sum_{k=2}^{30} \binom{k}{2}^2
\end{equation}
But it would be simpler to consider three cases.
$\textbf{(a)}$  The events $F$(fencing), $J$ (javelin) and $C$ (celebration) take up altogether $5$ days.
$\textbf{(b)}$ The events take up $4$ days.
$\textbf{(c)}$  The events take up $3$ days.
In case $\textbf{(a)}$, we have $\binom{31}{5}$ choices for the days of the events. But we have to multiply this binomial coefficient by $\binom{4}{2}$ to achieve the counting. And I don't understand why.
Could someone please explain the combinatoric signification of this multiplicative factor $\binom{4}{2}$? To what choices does it correspond?
I thank you in advance.
 A: Once the $5$ days are chosen, the last one must be for the celebration.

Of the other $4$ days, choose $2$ of them for fencing, so ${\large{\binom{4}{2}}}=6\;$choices.

Once the two fencing days are selected, the javelin days are forced.

Hence, since there are ${\large{\binom{31}{5}}}$ choices for the $5$ days, and for each such choice, there are ${\large{\binom{4}{2}}}$ ways to schedule the events within those $5$ days, the multiplication rule yields ${\large{\binom{31}{5}}}{\,\cdot\,}{\large{\binom{4}{2}}}$ ways to schedule the events, assuming the $5$-day scenario. 
A: In case $\textbf{(a)}$, we don't have to consider double contest days.
In case $\textbf{(b)}$, we have $\binom{31}{4}$ choices, the last one being reserved for the celebration day.
Then we have to choose $1$ day upon the $3$ remaining for the double contest day (where the events $F$ and $J$ are to be held simultaneously) !
Then there is two ways of allocating $F$ (or $J$) for the remaining $2$ days.
So in case $\textbf{(b)}$ we have $6\binom{31}{4}$ possibilities.
