Let's think about what the inequality is saying. If we multiply the left side out, we get $\frac{a_1 + 1}{2} \cdot \frac{a_2 + 2}{2} \cdot \frac{a_3 + 3}{2} \cdot \frac{a_4 + 4}{2} \cdot \frac{a_5 + 5}{2} \cdot \frac{a_6 + 6}{2} = \frac{(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)}{64}$. So the inequality is saying that the product of the numbers $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)$ is greater than $6! * 64$. Right, and since $64 \cdot 6! = 64 \cdot 720 = 4480$, the inequality is just saying that $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 4480$. So we just need to find the number of permutations of $(1,2,3,4,5,6)$ such that $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 4480$. Right. Let's think about this. If we want to make $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)$ as small as possible, we would want to let $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = 4$, $a_5 = 5$, and $a_6 = 6$. Oh yeah, since $64 \cdot 6! = 64 \cdot 720 = 46080$. Right. So the smallest possible value of $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)$ is $46080$. We know that $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 4480$, but we want to show that $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 46080$. Right. We need to show that all other permutations will satisfy this inequality. So let's assume that some permutation does not satisfy this inequality. Then we must have $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) \leq 46080$. Because of our assumption, we know that $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)$ is minimized. So we want to show that if we switch $a_j$ and $a_k$, then we will have $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 46080$. All we need to show is that $(a_j + j)(a_k + k) > (a_k + j)(a_j + k)$. Right. This is equivalent to $a_j a_k + j a_k + k a_j > a_j a_k + j a_j + k a_k$. And this is equivalent to $a_j k + a_k j > a_j j + a_k k$. Right, and if we subtract $a_j j + a_k k$ from both sides, we get $a_j k + a_k j - a_j j - a_k k > 0$, and if we factor, we get $a_j (k - j) + a_k (j - k) > 0$. And since $j < k$, we know that $k - j > 0$ and $j - k < 0$, so $a_j (k - j) + a_k (j - k) > 0$ is equivalent to $a_j (k - j) > a_k (k - j)$. So we have shown that if we switch $a_j$ and $a_k$, then we will have $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6) > 46080$. So we know that the permutation $(1,2,3,4,5,6)$ gives us the smallest value of $(a_1 + 1)(a_2 + 2)(a_3 + 3)(a_4 + 4)(a_5 + 5)(a_6 + 6)$. And since $(1,2,3,4,5,6)$ is just one permutation of $(1,2,3,4,5,6)$, we know that all other permutations will satisfy the inequality. So we have shown that there is only one permutation that does not satisfy the inequality. So there must be $720 - 1 = 719$ permutations that satisfy the inequality.


Find the number of permutations $(a_1, a_2, a_3, a_4, a_5, a_6)$ of $(1,2,3,4,5,6)$ that satisfy
\[\frac{a_1 + 1}{2} \cdot \frac{a_2 + 2}{2} \cdot \frac{a_3 + 3}{2} \cdot \frac{a_4 + 4}{2} \cdot \frac{a_5 + 5}{2} \cdot \frac{a_6 + 6}{2} > 6!.\]