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The drama club is holding auditions for 5 different parts in a one-act play. If 9 people audition, how many ways can the roles be assigned?

User MikeKulls
by
5.3k points

2 Answers

5 votes

Answer:

15,120

Explanation:

Formula: nPr= n!/(n-r)!

nPr= 9!/(9-5)! = 9!/4! = 15,120

The drama club is holding auditions for 5 different parts in a one-act play. If 9 people-example-1
User Rasel Khan
by
5.2k points
1 vote

Answer:

84 ways

Explanation:

Given

Total individuals = 9

Available roles = 5

Required

Number of ways roles can be assigned.

First, it'll be assumed that no two individual shares a role and no single individual takes two roles.

Having said this, the number of ways the roles can be assigned can be calculated using the combination formula; as follows.

ⁿCr = C(n,r)

Where n represents the total individuals

n = 9

r represents the available roles

r = 5

So,

ⁿCr = C(n,r) =>

⁹C₆ = C(9,6)

⁹C₆= 9!/((9-6)!6!)

⁹C₆ = 9!/(3!6!)

⁹C₆ = (9 * 8 * 7 * 6!)/(6! * 3 * 2 * 1)

⁹C₆ = (9 * 8 * 7)/(3 * 2 * 1)

⁹C₆ = 504/6

⁹C₆ = 84

Hence, 9 roles can be assigned to 5 people in 84 ways

User Gargkshitiz
by
5.6k points