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irinina [24]
1 year ago
5

Suppose we want to choose 3 letters, without replacement, from the 4 letters A, B, C, and D. How many ways can this be done, if

the order of the choices is relevant? how many ways can this be done, if the order of the choices is not relevant?
Mathematics
1 answer:
Vinvika [58]1 year ago
8 0

The number of ways can this be done, if the order of the choices is relevant will be 24. And the number of ways can this be done if the order of the choices is not relevant will be 4.

<h3>What are permutation and combination?</h3>

A permutation is an act of arranging items or elements in the correct order. Combinations are a way of selecting items or pieces from a group of objects or sets when the order of the components is immaterial.

Suppose we want to choose 3 letters, without replacement, from the 4 letters A, B, C, and D.

The number of ways can this be done, if the order of the choices is relevant will be

⁴P₃ = 4! / (4 - 3)!

⁴P₃ = 4 x 3 x 2 x 1 / 1

⁴P₃ = 24

The number of ways can this be done if the order of the choices is not relevant will be

⁴C₃ = 4! / [(4 - 3)! x 3!]

⁴C₃ = (4 x 3!) / (3! x 1)

⁴C₃ = 4

More about the permutation and the combination link is given below.

brainly.com/question/11732255

#SPJ1

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Lesechka [4]

Answer:

Option A.

Step-by-step explanation:

From the figure attached,

Given : ΔABC ~ ΔDEC

By the property of similarity,

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\frac{\text{AB}}{ED}=\frac{BC}{EC}=\frac{AC}{DC}

Since, \frac{\text{AB}}{ED}=\frac{AC}{DC}

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6x + 42x = 1344

48x = 1344

x = \frac{1344}{48}

x = 28 units

Therefore, Option (A). x = 28 units will be the answer.

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