Problem 7 · Five women and four men in a row
The alternation is forced: $5! \cdot 4!$.
Integer answer, at most 4 digitsFive women ($A$, $B$, $C$, $D$, $E$) and four men ($R$, $S$, $T$, $U$) must occupy nine seats in a row so that no woman sits next to another woman and no man sits next to another man. In how many ways can the seats be occupied? (Two ways are different if some seats are occupied by different people.)
Copa Cangur · SCM
Medium
Closed answer
Reasoned solution
Key idea: the condition forces the genders to alternate, and with $5$ women and $4$ men only one pattern fits.
If no two neighbours can share a gender, the row alternates WMWM… With $5$ women and $4$ men, the only pattern that fits is
$$W\;M\;W\;M\;W\;M\;W\;M\;W.$$
The women fill the $5$ seats marked $W$ in $5!$ ways and the men the $4$ seats marked $M$ in $4!$ ways:
$$5! \cdot 4! = 120 \cdot 24 = 2880.$$
Answer: 2880