Evaluate binomial coefficient expressions for a given value of k

Expert intro: This is a straightforward substitution problem, but the notation can look more intimidating than it really is.

Detailed walkthrough

We are given $a=\binom{k}{2},\qquad b=\binom{3}{k+1}.$

For $k=1$: $a=\binom{1}{2}=0,$ because there are no ways to choose 2 items from 1 item.

Also, $b=\binom{3}{1+1}=\binom{3}{2}=3.$

Therefore, $b-a=3-0=\boxed{3}.$

The answer is small, but the reasoning matters: binomial coefficients with an upper number smaller than the lower number are zero.

💡 Pitfall guide

Don’t confuse $\binom{3}{k+1}$ with $\binom{k+1}{3}$. The order matters. Also, remember that $\binom{1}{2}$ is not 1; it is 0 because the selection is impossible.

🔄 Real-world variant

If $k=2$, then $a=\binom{2}{2}=1$ and $b=\binom{3}{3}=1$, so $b-a=0$. If the question instead asked for $a+b$, you would simply add the two evaluated coefficients after substitution.

🔍 Related terms

binomial coefficient, combinatorics, substitution