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ValentinkaMS [17]
3 years ago
11

How many derangements of {1, 2, 3, 4, 5, 6} begin with the integers 1, 2, and 3, in some order?

Mathematics
1 answer:
iragen [17]3 years ago
8 0

Answer:

Total derangement = 4

Step-by-step explanation:

To find - How many derangement of {1, 2, 3, 4, 5, 6} begin with the integers 1, 2, and 3, in some order?

Solution -

Given that,

The first three numbers are 1, 2, 3

The only derangement possible are - 231 and 312

Now,

The remaining elements 4, 5, 6 are the 4th, 5th and 6th number in the derangement

Now,

In order for the permutation of the derangement, the only possibility for 123  are 564 and 645

∴ we get

The Total derangement are -

231564

231645

312564

312645

So,

Total derangement = 4

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Answer:

No

Step-by-step explanation:

The sequence is not an arithmetic sequence. For a sequence to be arithmetic, the difference between consecutive terms is a constant number which is termed as the common difference.

This means that the difference between the second term and the first term must be equal to the difference between the third term and the second term.

In the sequence above, the first term is 10. The difference between the first and second term is 15. When the first term is subtracted from the second, what we get is 5.

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3 years ago
10 - |3n - 2| = -9<br><br>please help :D​
lapo4ka [179]

Answer:

n =7 or n = -17/3

Step-by-step explanation:

10 - |3n - 2| = -9  (subtract 10 from both sides)

- |3n - 2| = -9 - 10

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|3n - 2| = 19

(3n - 2) = 19 or (3n - 2) = -19

For (3n - 2) = 19,

3n - 2 = 19

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For (3n - 2) = -19,

3n - 2 = -19

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Read more on Brainly.com - brainly.com/question/11678436#readmore


Step-by-step explanation


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5 0
2 years ago
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Which function is odd?
ddd [48]
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example
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