$100$ students can get grades that are divisible by 5 from $0$ to $100$. How many ways are there to assign their grades such that the average is at least 60? The solution should use inclusion-exclusion and can be left with sums.
I'm unsure of how to start with this one. Not even sure what to include-exclude. Any help would be appreciated.
Preparations: First, we formalize the problem. Let $\mathcal G=\{g\in 5\mathbb Z:0\le g\le 100\}$ be the grades. Then we want to know the size $|\mathcal S|$ of $$\mathcal S=\left\{g\in\mathcal G^{100}:\frac{1}{100}\sum_{i=1}^{100}g_i\ge 60\right\}=\left\{g\in\mathcal G^{100}:\sum_{i=1}^{100}g_i\ge 6000\right\}.$$ Now, let $\mathcal G_\circ=\{0,\dots,20\}$ and $\mathcal S_\circ=\{g\in\mathcal G_\circ^{100}:\sum_{i=1}^{100}g_i\ge 1200\}$. Then we have $\mathcal G=5\mathcal G_\circ$ and $$\mathcal S=\left\{5g:g\in\mathcal G_\circ^{100},5\sum_{i=1}^{100}g_i\ge 6000\right\}=5\mathcal S_\circ.$$ So, we have $|\mathcal S|=|\mathcal S_\circ|$ and don't have to deal with multiples of five and averages anymore.
Next, we split the set $\mathcal S_\circ=\bigcup_{t=1200}^{2000}\mathcal S_{t}$ into sets $\mathcal S_t=\{g\in\mathcal G_\circ^{100}:\sum_{i=1}^{100}g_i=t\}$ with fixed totals. This gives $|\mathcal S_\circ|=\sum_{t=1200}^{2000}|\mathcal S_t|$, and now it is sufficient to understand $\mathcal X=\mathcal S_t$ for fixed $1200\le t\le 2000$.
Counting: We follow the approach here and here. So, let $\mathcal T_s=\{x\in\mathbb Z_{\ge 0}^{100}:\sum_{i=1}^{100}x_i=t\}$ and let $\mathcal T_{s,i,u}=\{x\in\mathcal T_{s}:x_i\ge u\}$, which we only need for $\mathcal A_i=\mathcal T_{t,i,21}$ here. Then we have $\mathcal X=\mathcal T_{t}\setminus\bigcup_i\mathcal A_i$, so we can use the Inclusion-Exclusion-Principle to obtain $$|\mathcal X|=|\mathcal T_t|-\sum_{k=1}^{100}(-1)^{k+1}\sum_{1\le i_1<\cdots<i_k\le 100}\left|\bigcap_{j=1}^k\mathcal A_{i_j}\right|.$$ First, notice that $|\mathcal T_s|=\binom{s+100-1}{s}=\binom{s+99}{s}$ using stars and bars. Further, notice that $$\mathcal T_{s,i,u} =\left\{x\in\mathbb Z_{\ge 0}^{100}:\sum_{i=1}^{100}x_i=s,x_i\ge u\right\} =\left\{x+u\mathrm e_i:x\in\mathcal T_{s-u}\right\},$$ where $\mathrm e_i\in\{0,1\}^{100}$ denotes the $i$-th unit vector. Similarly, we have $$\bigcap_{j=1}^k\mathcal T_{s,i_j,u_j} =\left\{x\in\mathbb Z_{\ge 0}^{100}:\sum_{i=1}^{100}x_i=s,(x_{i_j})_j\ge (u_{j})_j\right\} =\left\{x+\sum_{j=1}^ku_j\mathrm e_{i_j}:x\in\mathcal T_{s-\sum_ju_j}\right\},$$ and hence $|\bigcap_{j=1}^k\mathcal T_{s,i_j,u_j}|=|\mathcal T_{s-\sum_ju_j}|=\binom{s-\sum_ju_j+99}{s-\sum_ju_j}$. With these results we obtain \begin{align*} |\mathcal X|&=\binom{t+99}{t}-\sum_{k=1}^{100}(-1)^{k+1}\sum_{1\le i_1<\cdots<i_k\le 100}\binom{t-21k+99}{t-21k}\\ &=\binom{t+99}{t}-\sum_{k=1}^{100}(-1)^{k+1}\binom{100}{k}\binom{t-21k+99}{t-21k}\\ &=\sum_{k=0}^{100}(-1)^{k}\binom{100}{k}\binom{t-21k+99}{t-21k}. \end{align*}
Alternative: We can also follow the approach here. The generating function for a single student is $f(x)=\sum_{g=0}^{20}x^g=\frac{1-x^{21}}{1-x}$ and the generating function for the sum is $S(x)=f(x)^{100}=\frac{1}{(1-x)^{100}}\sum_{s=0}^{100}\binom{100}{s}(-1)^sx^{21s}$. We recall the generating function $\frac{1}{(1-x)^{100}}=\sum_{n=0}^{\infty}\binom{n+100-1}{n}x^n$ for stars and bars from Wikipedia, and hence obtain $S(x)=\sum_{n=0}^\infty\sum_{s=0}^{100}\binom{n+100-1}{n}\binom{100}{s}(-1)^sx^{n+21s}$. Extracting the coefficient for $x^t$ yields $\sum_{k=0}^{100}(-1)^k\binom{t-21k+99}{t-21k}\binom{100}{k}$.
