How many $10$-digit numbers use only the digits $0, 1, 2$ with each digit appearing at least twice or not at all?

I know I need the coefficient of $\frac{x^{10}}{10!}$ in:

$$\left(1+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots \right)^3=(e^x-x)^3$$

but what do I do here?

  • $\begingroup$ Expand the RHS. $\endgroup$ – David May 15 '15 at 6:04
  • $\begingroup$ I expanded it but I don't see what to do from here. I don't see how to get an x^10 term when my highest term I have is x^3. $\endgroup$ – JackHallam May 15 '15 at 6:16
  • $\begingroup$ What did you get when you expanded the RHS? $\endgroup$ – David May 15 '15 at 6:21
  • $\begingroup$ i.gyazo.com/1791ce24af112318199c6d8bdcdfc8d8.png $\endgroup$ – JackHallam May 15 '15 at 6:22

You correctly expanded the RHS as $$(e^x)^3-3x(e^x)^2+3e^xx^2-x^3$$

Now write this as $$(e^{3x})-3x(e^{2x})+3e^xx^2-x^3$$

We look to the Taylor series of $e^{3x}$, found by substituting $y=3x$ in the Taylor series of $e^x$:

$$e^{3x} = 1+3x+\frac{3^2x^2}{2!} + \frac{3^3x^3}{3!} + \cdots + \frac{3^{10}x^{10}}{10!} + \cdots $$

Can you find those form $e^{2x}$ and $e^{x}$ yourself? If not, just leave a comment and I'll help you.

  • 1
    $\begingroup$ It took a little bit of algebra, but I got it. Thanks! $\endgroup$ – JackHallam May 15 '15 at 15:40
  • 2
    $\begingroup$ I wonder if the question implies that the first digit cannot be a 0 (since it asks for the number of 10 digit NUMBERS). In this case I think we need the coefficient of x^9/9! in 2*(exp(x) - 1)*(exp(x) - x)^2. which is 29306. $\endgroup$ – Geoffrey Critzer May 16 '15 at 20:00

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