Question

A catering service offers 12 appetizers, 9 main courses, and 7 desserts. A banquet chairperson is to select 8 appetizers, 8 main courses, and 6 desserts for a banquet. In how many ways can this be done?

Answer 1

$12-appetizers, \ 9- main\ courses, \ 7\ desserts\\ \\ selection:\ 8\ appetizers,\ 8\ main\ courses, 6\ desserts\\\\a=the\ number\ of\ selection\ of\ appetizers:\\ \\{12 \choose 8}= \frac{12!}{8!\cdot (12-8)!} = \frac{12\cdot11\cdot10\cdot9\cdot8!}{8!\cdot4\cdot3\cdot2} =11\cdot5\cdot9=495\\ \\c=the\ number\ of\ selection\ of\ main\ courses:\\ \\{9 \choose 8}= \frac{9!}{8!\cdot (9-8)!} = \frac{9\cdot8!}{8!\cdot 1} =9$

$d=the\ number\ of\ selection\ of\ desserts:\\ \\{7 \choose 6}= \frac{7!}{6!\cdot (7-6)!} = \frac{7\cdot6!}{6!\cdot 1} =7\\ \\the\ number\ of\ selection\ sets\ the\ banquet:\\ \\a\cdot c\cdot d=495\cdot9\cdot7=31185$

Answer 2

[(12! / (8!*4!) ]* [9! / (8!*1!) ] * [ 7!/ (6!*1!)] = ( 12 * 11 * 10 * 9 / 4 * 3 * 2 * 1) * 9 * 7 = 45 * 11 * 9 * 7 = 31185 ways