I have a question on Binomial theorem that goes like as the following:
Math 240 professors McBrain and DeCerveau are looking for problems for the weekly quiz. They found all four problems yesterday in their Combinatorial Grimoire, which contains 47 problems, but they can't remember which ones. McBrain chose the highest-numbered problem. DeCerveau picked the remaining problems. They discuss the number of possibilities: they agree if McBrain chose problem j for some $j\in[47]$ then DeCerveau had $\binom{j-1}3$ ways to complete the quiz. How many possibilities for the quiz are there?
Since we need to choose at least 3 questions out $j-1$, j is at least 4. I have come up with a general formula $\binom{47}j \binom j 1 \binom{j-1}3$ to say that there are $\binom{47}j$ to choose a $j$-elements subsets out [47], and we have for each subset $\binom j 1 \binom{j-1}3$ ways to choose 4 quiz questions.
I have tried to sum up all the possibilities as $$\binom{47}4 \binom 4 1 \binom{4-1}3 + \binom{47}5 \binom 5 1 \binom{5-1}3 + \cdots+\binom{47}{47} \binom{47}1 \binom{46}3,$$ but I am not able to come up with a concrete number as the total sum of possibilities.
Any help is appreciated!
The desired sum is $\sum_{j=4}^{47} \binom{j-1}{3}$, which, by the hockey-stick identity, is $\binom{47}{4}=178365$. This count is expected because you know they chose $4$ questions out of $47$.