A team has 10 members of different heights. Howmany ways are there forthe members to line up in two rows of five people so that each member in the back row is taller than the one imediately in front?

Guest Dec 18, 2018

#1**+2 **

\(\text{Split the team up into 2 subteams by height}\\ \text{The tallest 5 is one subteam, the shortest 5 the other}\\ \text{Any member of the tall team can stand in the back}\\ \text{Any member of the short team can stand in front}\\ \text{There are }5! \text{ ways for the 5 tall team members to arrange themselves}\\ \text{There are also }5! \text{ arrangements for the short team members}\\ \text{Thus there are }5! \times 5! = 120^2 = 14400 \text{ different arrangements}\)

.Rom Dec 18, 2018

#1**+2 **

Best Answer

\(\text{Split the team up into 2 subteams by height}\\ \text{The tallest 5 is one subteam, the shortest 5 the other}\\ \text{Any member of the tall team can stand in the back}\\ \text{Any member of the short team can stand in front}\\ \text{There are }5! \text{ ways for the 5 tall team members to arrange themselves}\\ \text{There are also }5! \text{ arrangements for the short team members}\\ \text{Thus there are }5! \times 5! = 120^2 = 14400 \text{ different arrangements}\)

Rom Dec 18, 2018