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Andre45 [30]
3 years ago
10

How many partitions of $12$ are there that have at least four parts, such that the largest, second-largest, third-largest, and f

ourth-largest parts are respectively greater than or equal to $4,3,2,1$?
Mathematics
1 answer:
leonid [27]3 years ago
6 0

Answer:

4 , 4 , 2 , 2

4, 3 , 3 , 2

Step-by-step explanation:

Given data in the problem:

total value = 12

now,

each partition should be greater than or equal to 4 , 3 , 2 , 1

now, the sum of the above terms comes as:

4 + 3 + 2 + 1 = 10

difference from the original values : 12 - 10 = 2

so we can divide 2 into two equal parts to the number 3 and 1

thus, we get the partitions as:

4 , (3 + 1) , 2 , (1 + 1) = 4 , 4 , 2 , 2

and another partition can be

4, 3, (2 + 1), (1 + 1) = 4, 3 , 3 , 2

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Step-by-step explanation:

Answer:

0.430625 = 0.431

Step-by-step explanation:

Let W represent winning, D represent a draw and L represent a loss.

12+ points can be garnered in each of the following ways.

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The probability of getting 12+ points is the sum of all these 7 probabilities.

Knowing that P(W) = 0.5

P(D) = 0.1

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P(6W 0D 0L) = [6!/(6!0!0!)] 0.5⁶ 0.1⁰ 0.4⁰ = 0.015625

P(5W 1D 0L) = [6!/(5!1!0!)] 0.5⁵ 0.1¹ 0.4⁰ = 0.01875

P(5W 0D 1L) = [6!/(5!0!1!)] 0.5⁵ 0.1⁰ 0.4¹ = 0.075

P(4W 2D 0L) = [6!/(4!2!0!)] 0.5⁴ 0.1² 0.4⁰ = 0.09375

P(4W 1D 1L) = [6!/(4!1!1!)] 0.5⁴ 0.1¹ 0.4¹ = 0.075

P(4W 0D 2L) = [6!/(4!0!2!)] 0.5⁴ 0.1⁰ 0.4² = 0.15

P(3W 3D 0L) = [6!/(3!3!0!)] 0.5³ 0.1³ 0.4⁰ = 0.0025

The probability of getting 12+ points = 0.015625 + 0.01875 + 0.075 + 0.09375 + 0.075 + 0.15 + 0.0025 = 0.430625

Read more on Brainly.com - brainly.com/question/14850440#readmore

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