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

Question

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?

Answer 1

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.

Answer 2

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?

Answer 3

$C_r^{16}=C_{r+2}^{16}$ Then using the definition of combination you'll get

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