If 3030-r+33031-r+33032-r+3033-r=mr for 3 r30, then m equals :

Question

If $\displaystyle\binom{30}{30-r}+3\binom{30}{31-r}+3\binom{30}{32-r}+\binom{30}{33-r}=\binom{m}{r}$ for $3\le r\le30$, then $m$ equals :

Accepted Answer

Using symmetry $\binom{30}{k}=\binom{30}{30-k}$:

$$\binom{30}{30-r}=\binom{30}{r},\quad \binom{30}{31-r}=\binom{30}{r-1},\quad \binom{30}{32-r}=\binom{30}{r-2},\quad\binom{30}{33-r}=\binom{30}{r-3}$$

So the left side becomes $\binom{30}{r}+3\binom{30}{r-1}+3\binom{30}{r-2}+\binom{30}{r-3}$, whose coefficients $1,3,3,1$ are those of $(1+x)^3$. Indeed,

$$(1+x)^{33}=(1+x)^{30}(1+x)^3=(1+x)^{30}(1+3x+3x^2+x^3)$$

Comparing the coefficient of $x^r$ on both sides (using Vandermonde-type expansion) gives

$$\binom{30}{r}+3\binom{30}{r-1}+3\binom{30}{r-2}+\binom{30}{r-3}=\binom{33}{r}$$

So $m=33$.

Solutions · 1