In a drama class of 18 students, nine are selected to be actors in a play,

Question

In a drama class of

18 students, nine are selected to be

actors in a play, five will build sets, and

four will be stage hands. In how many

ways could the class be divided up?

a) Make your calculations by selecting

the actors first.

b) Make your calculations by selecting

the set builders first.

c) Compare your answers. Explain the

results.

I don't understand the wording of the question, please explain for each scenario, followed by steps! Thank you

Answer 1

Here is what happens.

Case a: Nine out of eighteen participants are actors. Therefore the combinations (nine actors out of eighteen people) are 18! / (9! * 9!), that is eighteen factorial divided by nine factorial divided by nine factorial. To each such combination we must consider the number of combinations among the remaining nine non-actors, since five will be set builders and four will be stage hands. The number of the latter combinations is 9! / (5! * 4!) (nine factorial divided by five factorial divided by four factorial). The total number of these combinations is the product of these two numbers, which comes out to be:

18! / (9! * 5! * 4!)

Case b: We first start with the set builders, so the combinations for set builders (five out of eighteen) are: 18! / (13! * 5!). Out of the remaining thirteen non set builders, nine will be actors and four will be stage hands. The number of combinations for these thirteen is: 13! / (9! * 4*), and the total number of combinations is the product of these two numbers:

18! / (9! * 5! * 4!)

Case c: as expected, the answers from case a and case b agree.