1

I have this equation: $$\binom{39}{5+2x}=\binom{39}{2x-2}$$

And I don't know how to solve it. I've tried by the definition of combination but I get stuck.

I get stuck here:

StubbornAtom
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tobi
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  • The definition is a good start. Can you show where you get stuck? – mathreadler Oct 12 '16 at 13:21
  • If one would use the symmetric property in the answers below, one would need to show that it would yield all solutions. That is maybe a suitable exercise to do afterwards. – mathreadler Oct 12 '16 at 13:49

4 Answers4

5

Hint:

$$\binom np=\binom nq\iff p+q=n \text{ OR } p=q$$

ajotatxe
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we have

$\binom{n}{p}=\binom{n}{n-p}$

so your equation offers two possibilities:

$5+2x=2x-2$ which is not possible

or

$5+2x=39-(2x-2) $ which gives

$x=\frac{36}{4}=9$.

0

You can solve this question very easily by using the property $\binom{n}{r} = \binom{n}{n-r} $

Change the term on the RHS to$$ \binom{39}{41-2x} $$ Now equate the bases of both combinations and you'll get $$5+ 2x = 41-2x $$ Solving for x you'll get $$x = 9 $$

Tejus
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Think about $\binom{n}{k} = \binom{n}{n-k}$.

Then

$39-(5+2x)=2x-2$

$39-5-2x=2x-2$

$34-2x=2x-2$

$4x=36$

$x=9$

Alternatively you could have set it up also as

$5+2x=39-(2x-2)$

and solved for $x$ as well.