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Oh yes, there is! Here: $\frac{n!}{r!(n-r)!}$- combinations; order does not matter

The order does matter( Permutations): $\frac{n!}{(n-r)!}$

If you take AoPS, they give you wonderful examples! 

Also, a quote from CPhill's explanation: "We have   C(16, 4)   = 1820   possible sets of any four of the points."

We can express this as $C\binom{16}{3}$ , where it is combinations. 

It's not a fraction. However, it means 16 choose 4; it computes the number of ways to choose items from a collection of items. 

Let $x$ be the cost of the food. Then $1.2(1.06x)=25.44\implies x=\boxed{\$20}$.

Another way to approach it is realization!

While we can solve for $a$ and $b$ individually, it is simpler to note that $f(1) = a + b$. Thus, substituting $1$ into the given equation, we obtain$g(f(1)) = 3 \cdot 1 + 4 = 7.$Thus, $g(f(1)) = 2 \cdot f(1) - 5 = 7 \Longrightarrow f(1) = \boxed{6}.$

Also, $e$ if you want to know is named after Euler. Read more information, here: https://www.mathsisfun.com/numbers/e-eulers-number.html

Welcome! 

Sorry if my explanation was confusing. Read more about the quadratic formula https://www.purplemath.com/modules/quadform.htm and https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-review, and I'm sure that you'll get a deeper understanding. 

Nope, not quite. The answer is actually, (C), $5069.96. First, you multiply $\:0.045\cdot \:7$ , which is $0.315$ . Now, $e^{0.315}=1.37026\dots$, so we have $3700\cdot \:1.37026\dots =5069.95945\dots \approx=5069.96$ or (C). 

Anytime! Glad to help out users!