Larger or smaller?

Bk Lim's solution is perfect. I'm including mine because it's only slightly differently stated.

In both (a) and (b), you have to select 5 distinct numbers out of 10 possibilities and put them into the only allowed order. This is just $\begin{pmatrix} 10 \\ 5 \end{pmatrix} = 252$ .

In case (a) only, you have to ignore the selections for which the left-most digit is zero, since this is actually a 4-digit number. This is equivalent to subtracting $\begin{pmatrix} 9 \\ 4 \end{pmatrix} = 126$ .

Therefore:

$a = \begin{pmatrix} 10 \\ 5 \end{pmatrix} - \begin{pmatrix} 9 \\ 4 \end{pmatrix} = 252 - 126 = 126$

and:

$b = \begin{pmatrix} 10 \\ 5 \end{pmatrix}= 252$

Add these together:

$a + b = 252 + 126 = \boxed{378}$

$a = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{5!}$ We are picking 5 numbers out of 9, only $\frac{1}{5!}$ fulfill the requirement .

$b = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5!}$ We are picking 5 numbers out of 10, only $\frac{1}{5!}$ fulfill the requirement .

$a+b =\frac{15 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5!} = 378$