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2 years ago

10

A theater company with 9 actors is putting on a play that has 3 roles. In how many different ways can the director assign actors

to those roles?

2 answers:

steposvetlana [31]2 years ago

6 0

9!/(3!*6!)=7*8*9/6=84

Cerrena [4.2K]2 years ago

5 0

He has 9 choices for the first role, and that leaves him 8 people to choose for the second so he has 9*8 = 72 choices for the first two roles; that leaves 7 possible choices for the third role so that gives him 9*8*7 choices or 504 different ways he can assign the roles. i think this is correct.

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3 years ago

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Norma-Jean [14]

Answer:

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Step-by-step explanation:

Given

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See comment for missing part of the question

Required

Complete the expression to determine the dimension of a rectangle

We have:

$x(x-3) =10$

Open bracket

$x^2 -3x = 10$

Equate to 0

$x^2 -3x - 10 =0$

Expand

$x^2 + 2x - 5x - 10 = 0$

Factorize

$x(x + 2) - 5(x + 2) = 0$

Factor out x + 2

$(x - 5)(x + 2) = 0$

Solve for x

$x - 5 = 0$ or $x + 2 = 0$

$x = 5$ or $x = -2$

The value of x cannot be negative

So:

$x = 5$

Recall that:

$Length = x$

$Width = x - 3$

So:

$Length =5$

$Width = 2$ ---- i.e. 5 - 3

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2 years ago