Question

Use the Fundamental Counting Principle to solve.

Answer 1

$number \: of \: performances = 9 \\ note = \\ one \: insists \: on \: being \: last \\ number \: of \: flexible \: performances = 9 - 1 = 8 \\ \\ c( \gamma ) = 8P8 \times 1 \\ c( \gamma ) = 8! \\ c( \gamma ) = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ c( \gamma ) = 40320 \: ways$