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sasho [114]
1 year ago
10

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]1 year 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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To get the solution, we are looking for, we need to point out what we know.

1. We assume, that the number 120 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 120, so we can write it down as 100%=120.
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now we have:
371 is 309.166666667% of 120
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