MathProblemSolver101

(a) $10 \cdot 9 \cdot 8 = \boxed{720}$

(b) $10^3 = \boxed{1000}$

"The proof is difficult to explain concisely and without a diagram."

This reminds me of the quote "I have discovered a truly remarkable proof of this theorem which this margin is too small to contain"

Your question is vague. What types of questions? Like, how to get started? How to use states? Complementary? etc.

Your question is like: "How do I be better at basketball?" A good question is something like: "How do I become better at 3 pointers and free throws?"

One possible improvement to your question is asking a particular aspect, or topic, of probability.

Let $p(x) = x^2, q(x) = 2x^3+1,$ and $r(x) = 7x^4 - 5x^3 - 6x - 2.$ Find $p(q(x)) + r(x).$

We have $(2x^3 + 1)^2 + (7x^4 - 5x^3 - 6x - 2) = 2x^6 + 4x^3 + 1 + 7x^4 - 5x^3 - 6x - 2 = \boxed{2x^6 + 7x^4 - x^3 -6x - 1}.$

Wow that's slick

We try a complementary approach

Multiples of 3 from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{3}}\right \rfloor = \left \lfloor{\frac{1200}{3}} \right \rfloor = 400$

Multiples of 4 from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{4}}\right \rfloor = \left \lfloor{\frac{1200}{4}} \right \rfloor = 300$

Multiples of $4 \cdot 3 = 12$ from 1 - 1200: $\left \lfloor{\frac{1200 - 1 + 1}{12}}\right \rfloor = \left \lfloor{\frac{1200}{12}} \right \rfloor = 100$

$1200 - (400 + 300 - 100) = 1200 - 600 = \boxed{600}$

8 letters total

3 I's

2 N's

1 of everything else

$\frac{8!}{3!2!} = 4 \cdot 7 \cdot 6 \cdot 5 \cdot 4 = 20 \cdot 24 \cdot 7 = 140 \cdot 24 = \boxed{3360}$

Says certificate is valid for me... no popup shows- I tried this with every browser (edge,safari,chrome,chromium,firefox) and it does not show up. On a 2017 MacOS High Sierra 10.13.4

By the Fundamental Theorem of Lack of Information, this problem becomes trival: no solutions.