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If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.

Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?

Tianlalu
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Gold Pony Boy
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3 Answers3

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Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?

Carl Schildkraut
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Note that $\binom{n}{k} = \binom{n}{n-k}.$

What equation does this give for $r$ in your scenario? What is the solution to that equation?

Note that $\binom{n}{a+1} > \binom{n}{a}$ when $a < \lfloor n/2 \rfloor,$ hence an integer can be taken on at most twice as you range among $\binom{n}{0}, \dots, \binom{n}{n}.$ Since $r \ne r+2,$ the value of $r$ that you find is unique.

Display name
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  • Your equation does not imply $$\binom na=\binom nb\implies a=n-b$$ – Kemono Chen Dec 03 '18 at 04:52
  • @KemonoChen Note that $\binom{n}{a+1} > \binom{n}{a}$ when $a < \lfloor n/2 \rfloor,$ hence an integer can be taken on at most twice as you range among $\binom{n}{0}, \dots, \binom{n}{n}.$ – Display name Dec 03 '18 at 05:00
  • Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect. – Kemono Chen Dec 03 '18 at 05:01
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$C_r^{16}=C_{r+2}^{16}$ Then using the definition of combination you'll get

$$\frac{16!}{r!(16-r)!}=\frac{16!}{(r+2)!(16-r-2)!}$$ $$\implies \frac{1}{r!(16-r)(15-r).(14-r)!}=\frac{1}{(r+2)(r+1).r!.(14-r)!}$$ $$\implies (16-r)(15-r)=(r+2)(r+1)$$ $$\implies 240-31r+r^2=r^2+3r+2$$ $$\implies 34r=238$$ $$\implies r=7$$

Fareed Abi Farraj
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  • Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7 – Gold Pony Boy Dec 03 '18 at 05:06
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    @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation. – mathematics2x2life Dec 03 '18 at 05:07
  • @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it. – Gold Pony Boy Dec 03 '18 at 05:12
  • @mathematics2x2life Actually wait,where did the r! from the left side go after the first step? – Gold Pony Boy Dec 03 '18 at 05:24
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    @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side – Fareed Abi Farraj Dec 03 '18 at 05:30
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    This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x – smci Dec 03 '18 at 17:46