Ans: B
The required number
$\begin{array}{cl}
= & C^{15}_1 \times C^{14}_4 \\
= & 15\ 015
\end{array}$
Resources for HKDSE Mathematics
Resources for HKDSE Mathematics
The required number
$\begin{array}{cl}
= & C^{15}_1 \times C^{14}_4 \\
= & 15\ 015
\end{array}$
