Why must we consider seemingly unnecessary cases for combinatorics problems? Consider the following question:
"A committee of 7 students must be selected from 4 tenth-grade students, 6 eleventh-grade students, and 8 twelfth-grade students. How many committees are possible if there must be at least 5 twelfth-grade students on the committee?"
My solution:
There are 8 twelfth-grade students, and at least five must be selected. Five will be selected from a group of 8, and the remaining 3 will be put into a separate group with the tenth and eleventh-graders, to form a group of (3+4+6) = 13 students. From this group, 2 will be selected, to form a committee of (2+5) = 7 students.
(₈C₅)(₁₃C₂) = 4368 possible committees.
Official solution:
"There are 8 twelfth-grade students, of which five, or six, or seven (but not eight because there are only 7 members of the committee), can be selected.
Case 1: Five twelfth-grade students are selected from a group of eight. The 10 remaining students from the other two grades are grouped together, and two of them are chosen to complete the committee.
(₈C₅)(₁₀C₂)
Case 2: Six twelfth-grade students are selected from a group of eight. The 10 remaining students from the other two grades are grouped together, and one of them is chosen to join the twelfth-graders in the committee.
(₈C₆)(₁₀C₁)
Case 3: Seven twelfth-grade students are selected from a group of eight. Seven students have been selected, and the committee is complete.
(₈C₇)(₁₀C₀)
Case 1 + Case 2 + Case 3 =
(₈C₅)(₁₀C₂) +
(₈C₆)(₁₀C₁) + (₈C₇)(₁₀C₀) = 2808 possible different committees."
I understand the approach to the official solution. However, it is unclear where the unselected twelfth-grade students disappear to. Why are they not included in the other group? In my solution, there is no need to consider cases, because the "at least five twelfth-grade students" condition is already satisfied. Five are guaranteed from the original group, and it is entirely possible that one or two more will join the committee, as they are included in the separate expression (₁₃C₂). I expected the two solutions to be equivalent, and reach the same integral solution. However, they did not. Surely there is a mathematical explanation for this.
 A: After having chosen five of the twelfth-graders, your solution subjects the remaining twelfth-graders to another round of selection, giving them a "second chance", thereby creating more possibilities than if you had carefully divided the entire selection process into three mutually exclusive cases.
A: By designating five of the seniors as the seniors as the seniors who serve on the committee and any additional seniors as additional members, you count each committee with more than five seniors multiple times.  In particular, you count each committee with six seniors $\binom{6}{5} = 6$ times, once for each of the six ways you could have designated five of those six seniors as the seniors on the committee and each committee with seven seniors $\binom{7}{5} = 21$ times, once for each of the $21$ ways you could have designated five of those seven seniors as the seniors on the committee.
To illustrate the case with six seniors, suppose that Andrew, Barbara, Cleo, Darren, Erin, Fiona, Gina, and Henry are the eight seniors and that Ines is a junior.  If Andrew, Barbara, Cleo, Darren, Erin, Fiona, and Ines are selected to serve on the committee, your method counts this particular committee six times.
$$
\begin{array}{l | l}
\text{five seniors} & \text{additional people}\\ \hline
\text{Andrew, Barbara, Cleo, Darren, Erin} & \text{Fiona, Ines}\\
\text{Andrew, Barbara, Cleo, Darren, Fiona} & \text{Erin, Ines}\\
\text{Andrew, Barbara, Cleo, Erin, Fiona} & \text{Darren, Ines}\\
\text{Andrew, Barbara, Darren, Erin, Fiona} & \text{Cleo, Ines}\\
\text{Andrew, Cleo, Darren, Erin, Fiona} & \text{Barbara, Ines}\\
\text{Barbara, Cleo, Darren, Erin, Fiona} & \text{Andrew, Ines}\\
\end{array}
$$
Similarly, if the seven seniors Barbara, Cleo, Darren, Erin, Fiona, Gina, and Henry are selected to serve on the committee, you can verify that there are $21$ ways to designate five of them as the five seniors and the other two as the additional people.
Notice that
$$\binom{8}{5}\binom{10}{2} + \color{red}{\binom{6}{5}}\binom{8}{6}\binom{10}{1} + \color{red}{\binom{7}{5}}\binom{8}{7}\binom{10}{0} = \color{red}{4368}$$
