Counting number of possible grade assignments Given 100 students who are to be assigned a grade 1 to 5, what is the number of possible assignments if we are interested in knowing what grade each student received. Furthermore, we are given the constraints that at least 30 students receive a grade of 4.
The way I approached this problem was by saying that since in all grade distributions at least 30 students must achieve a grade of 4 then we can "reserve" those using $100\choose 30$. This leaves us with 70 students that can be assigned grades using $5^{70}$. The final result would then be given by
$${100\choose 30}\cdot{5^{70}}$$
However, the actual solution is $$\sum_{k=30}^{100}{100\choose k}{\cdot}{4^{100-k}}$$
which also makes sense to me. I just can't seem to make out how exactly the two constructs differ and what my version is actually representing.
 A: let's use smaller numbers for intuition.  Assume two grade levels and 3 students.  Assume again at least one student gets lower grade.  So you have these scenarios
$\{1,2,2\}, \{2,1,2\}, \{2,2,1\}, \{1,1,2\},\{1,2,1\},\{2,1,1\}, \{1,1,1\}$
using your way will give $ {3 \choose 1} 2^2 = 12$, which is clearly wrong.
Using the proposed solution: 
$$\sum_{k=1}^{3}{3\choose k}{1^{3-k}}= 3 + 3 + 1 = 7$$
A: You solution is overcounting situations where more than $30$ people receive a grade of four. In your method, you select $30$ people to receive a grade of four, then assign the remaining grades arbitrarily. These initial $30$ people are given a special treatment they should not have.
Example: Consider the following two ways to assign grades according to your formula:


*

*
*

*Choose persons $1$ to $30$ to receive a grade of $4$.

*For everyone else: assign persons $31$ to $40$ a grade of $4$, and persons $41$ to $100$ a grade of $1$.


*
*

*Choose persons $11$ to $40$ to receive a grade of $4$.

*For everyone else: assign persons $1$ to $10$ a grade of $4$, and persons $41$ to $100$ a grade of $1$.



Both of these result in the same grade distribution (persons $1$ to $40$ have $4$, persons $41$ to $100$ have $1$), but they are counted as distinct by your formula. 
To eliminate this overcounting, you choose all the people who get grades of $4$ at once, and then assign the grades $1,2,3,5$ arbitrarily to the others. There are now only four choices for each other students. 

Your mistake is a very common pitfall for people learning combinatorics. Learning how to spot this mistake and avoid it in the future is a very important step for becoming good at combinatorics.
