probability

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probability

by sud21 » Sat Jan 28, 2012 10:21 pm
How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?

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by rijul007 » Sat Jan 28, 2012 11:05 pm
sud21 wrote:How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?
Total number of arrangements = 5!

Let us assume that 24 is a sigle element
Number of arrangements of 1, 24,3,5 = 4!
Nuumber of arrangemtns of 1, 42,3,5 = 4!

Number of arrangements such that 2,4 are not adjacent = 5! - 2*4! = 120 - 48 = 72

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by Scott@TargetTestPrep » Wed Jan 03, 2018 4:10 pm
sud21 wrote:How many arrangements of the digits 1, 2, 3, 4, and 5 are there, such that 2 and 4 are not adjacent?
We can use the following formula:

Total number of arrangements = # ways with 2 and 4 adjacent + # ways with 2 and 4 not adjacent

The total number of arrangements is 5! = 120

Next we can determine the number of arrangements when 2 and 4 are adjacent. We can make the numbers 2 and 4 one placeholder, such that there are 4 total positions or 4! = 24. However, we must include that we can arrange the 2 and 4 in 2! = 2 ways. So, the total number of ways is 24 x 2 = 48.

Therefore, the number of ways with 2 and 4 not adjacent is 120 - 48 = 72

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