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2 years ago

In how many ways can 7 persons be arranged in a line such that (a) two particular persons are never together.

Mathematics

1 answer:

ivanzaharov [21]2 years ago

3 0

Answer:

3600 ways

Step-by-step explanation:

person A has 7 places to choose from :

→ He has 2 places ,one to the extreme left of the line ,the other to the extreme right of the line

If he chose one of those two ,person B will have 5 choices and the other 5 persons will have 5! Choices.

⇒ number of arrangements = 2×5×5! = 1 200

→ But Person A also , can choose one of the 5 places in between the two extremes .

If he chose one of those 5 ,person B will have 4 choices and the other 5 persons wil have 5! Choices.

⇒ number of arrangements = 5×4×5! = 2 400

In Total they can be arranged in :

1200 + 2400 = 3600 ways

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Gwar [14]

We can use the Сosine formula to solve this problem.
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The smallest angle of the triangle lies opposite the smallest side, so we need to find m∠C.

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Now we can use Bradis's Table (I don't know the name in English, maybe Trigonometric Table?) to find m∠C:

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I hope this helps

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3 years ago

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Step-by-step explanation:

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Read 2 more answers

What are the solutions of the quadratic equation (x + 3)(x +3) = 49? Ax = -2 and x = -16 Bx = 2 and x = -10 Cx = 4 and x = -10 D

Ivahew [28]

Answer:

4 and -10

Step-by-step explanation:

$\displaystyle (x + 3)(x +3) = 49 \\ x^2+3x+3x+9=49 \\ x^2 +6x+9=49 \\ x^2 + 6x + 9 - 49 = 0 \\x^2+6x-40=0 \\\\ \Delta=b^2-4ac \\ \Delta=6^2-4 \cdot 1 \cdot (-40) \\ \Delta=36+160 \\ \Delta=196 \\ \\ X_{1,2}=\frac{-b \pm \sqrt{\Delta} }{2a} \\ \\ X_1=\frac{-b+\sqrt{\Delta} }{2a} = \frac{-6+14}{2} = \frac{8}{2}=4 \\ \\ X_2=\frac{-b-\sqrt{\Delta} }{2a} = \frac{-6-14}{2}=\frac{-20}{2} = -10$

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3 years ago