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3 years ago

13

Jing jing works in a store and she spends some of her time organizing the stationary section. If there are nine different items

on the rack, how many ways can she organize the rack so that the two most expensive items are not together?

1 answer:

Alexus [3.1K]3 years ago

4 0

Answer:

282240 ways

Step-by-step explanation:

number of items = 9 different items

Ranking based on worth

<u>determine number of ways to arrange the 9 items such that the 2 most expensive are apart</u>

first step: consider the 2expensive items as 1

∴ number of permutation = [ ( 9 - 2 ) + 1 ]! * 2!

= [ 8 ] ! * 2! = 80640

Total number of permutation/arrangement = 9! = 362880

hence to arrange the 9 items without having the two most expensive items together

= 362,880 - 80640

= 282240 ways

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kvv77 [185]

QUESTION 1

The given inequality is

$y\leq x-3$ and $y\geq -x-2$ .

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$-2\leq 3-3$ and $-2\geq -3-2$ .

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QUESTION 2

The given inequality is

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If (1,4) is a solution; then it must satisfy both inequalities.

We put x=1 and y=4 in to both inequalities.

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Both inequalities are satisfied, hence (1,4) is a solution to the given system of inequality.

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QUESTION 3

The given inequality is

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We put x=0 and y=-2 in to both inequalities.

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QUESTION 4

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3 years ago

Right answers plzzzz

Andrew [12]

D.

a.

b.

hope that helped

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