# Evaluate the sum $22\binom{26}{0} + 21\binom{26}{1} + 20\binom{26}{2} + \cdots + (-3)\binom{26}{25} + (-4)\binom{26}{26}.$

Evaluate the sum $$22\binom{26}{0} + 21\binom{26}{1} + 20\binom{26}{2} + \cdots + (-3)\binom{26}{25} + (-4)\binom{26}{26}.$$

Is there an obvious shortcut I'm missing? I know I can't just do the whole calculation! That would take years! But I don't see another way. Solutions are greatly appreciated!

• $22=26-4$, $21=25-4$, .. can you proceed from here ? – G Cab Oct 7 '16 at 21:00

Hint: ${n \choose m}={n\choose n-m}$. Edit: Note that this should be easy to show both combinatorially and algebraically.
• I got $697585176,$ is that correct? – Dreamer Oct 7 '16 at 21:17
• @Regina Using my hint, we can rewrite this expression so that each binomial coefficient has a coefficient of $9$. This is because for $i<13$, we have $(22-i){26 \choose i}$ and $(-4+i){26 \choose {26-i}}$ which we can sum to $(18){26 \choose i}$, and then redistribute it again into $9{26 \choose i}+9{26 \choose 26-i}$ by using the hint again. For $i=13$, we have only the term $9{26\choose 13}$. This gives us $9\sum_{i=0}^{26} {26\choose i}$. What does that sum represent combinatorially if you ignore the $9$? – Kevin Long Oct 7 '16 at 21:52
• Is the answer $9 \cdot 2^{26}$? – Dreamer Oct 7 '16 at 22:00