Hi everyone, first time posting on this forum and was hoping if someone would be willing to explain how to do the below question. Any help appreciated, thanks.
The question is to find the number of solutions to $\displaystyle \sum\limits_{k = 1}^5 {u_k } = 24,\quad 4 \le u_1 \;\& \;u_2 < 7$
Let $\displaystyle S_k$ be the event that $\displaystyle u_1=k$.
Let $\displaystyle T_k$ be the event that $\displaystyle u_2=k$.
Let $\displaystyle \|S_k\|$ be the number of solutions in $\displaystyle S_k$.
The answer to the posed question is $\displaystyle \sum\limits_{k = 4}^6 {\sum\limits_{j = 4}^6 {\left\| {S_k \cap T_j } \right\|} }$,
where $\displaystyle \left\| {S_k \cap T_j } \right\|=\binom{26-k-j}{2}$.