RRB Paramedical Staff Nurse Model Paper 5

CASE I: MF, MF, MF, MF CASE II: FM, FM, FM, FM Let us arrange 4 Males in 4! ways, then 4 Females can be arranged in ${}_{}^4{\mathrm{P}}_4^{}$ ways, either left of Males or Right of Males.
Hence required numbers of arrangements are
$=4!{\times }^4{\mathit{P}}_4+4!\times {}^4{\mathit{P}}_4$
$=2\times 576$
$=1152$