Management Aptitude Test Sep 2011 Paper

(3) When first two digits of the number is 41, then remaining four digits can be arranged in ${}_{}^8{\mathit{P}}_4^{}$ ways.
Similarly, for 42, 46, 62, and 64, the remaining four digits can be arranged in ${}_{}^8{\mathit{P}}_4^{}$ ways.
Hence, total number of numbers having all six digits distinct = 5 x ${}_{}^8{\mathit{P}}_4^{}$
= 5 x $\frac{8!}{4!}$ = 5 x 8 x 7 x 6 x 5 = 8400