Combinations equation

Asked Apr 14 '23 at 14:11

Active Apr 14 '23 at 14:11

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Solve the equation

We can write it as follows: $$\dfrac{x!}{(x-1)!(x-x+1!)}+\dfrac{x!}{(x-2)!(x-x+2)!}+\dfrac{x!}{(x-3)!(x-x+3)!}+...+\\\dfrac{x!}{(x-8)!(x-x+8)!}+\dfrac{x!}{(x-9)!(x-x+9)!}+\dfrac{x!}{(x-10)!(x-x+10)!}=1023\\\iff\dfrac{x}{1!}+\dfrac{(x-1)x}{2!}+\dfrac{(x-2)(x-1)x}{3!}+...+\dfrac{(x-9)...(x-1)x}{10!}=1023$$ which doesn't seem very helpful. What's the approach?

asked Apr 14 '23 at 14:11

yinivem462

Adding $\binom{x}{x-0}$ to the LHS and $1$ to the RHS, the RHS becomes $2^{10}$, and $x=10$ is one possible solution. – peterwhy Apr 14 '23 at 14:20
Hint: If $m>n>k$ then $\binom mk>\binom nk.$ – Thomas Andrews Apr 14 '23 at 14:22
@peterwhy, how does the LHS become $2^{10}$? – yinivem462 Apr 14 '23 at 14:29
@ThomasAndrews, I really don't see how this can be applied in our problem. Could you further clarify for me, please? – yinivem462 Apr 14 '23 at 14:33
@yinivem462 For Peter's comment: $\binom x{x-a}=\binom xa,$ so binomial theorem says $$(1+1)^{10}=\binom{10}0+\binom{10}1+\cdots +\binom{10}{10},$$ so subtract $1=\binom{10}0$ from both sides. – Thomas Andrews Apr 14 '23 at 14:33

Combinations equation

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