It's easier to consider that the number you are looking for = (total arrangements of 1, 2, 3, 4 and 5) - (number of arrangements in which 2 and 4 are adjacent).
Now, the total number of arrangements will be 5! = 120.
The number of arrangements in which 2 and 4 are adjacent will be a bit harder to guess, IMHO. There are a few cases:
a. 2 and 4 are to the left: 24xxx and 42xxx makes 2*6 = 12 cases (since there are 6 ways of arranging 1, 3 and 5)
b. 2 and 4 are like so x24xx and x42xxx makes 2*6 = 12 cases
c. 2 and 4 are like so xx24x and xx42x will makes 2*6 = 12 cases
d. 2 and 4 are to the right xxx24 and xxx42 makes 2*6 = 12 cases
for a grand total of 48 cases.
This makes the number you're looking for 120 - 48 = 72.
Simple arrangement question
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Agreeing with DanaJ, but using a slightly different method which might save time for those comfortable with Combinations.
Total number of arrangements possible for 5 numbers = 5! = 120.
For finding total number of selections where 2 & 4 occur together, lets take 2&4 and bundle it to form one number X. So, total number of arrangements possible for 1,3,5 & x = 4! = 24.
But X is made up of 2 numbers 2 & 4, thus, number of ways X can be formed = 2! = 2.
So, total number of arrangements in which 2 & 4 are adjacent = 24 * 2 = 48.
Hence, total number of arrangements in which 2 & 4 are not adjacent = 120 - 48 = 72.
Total number of arrangements possible for 5 numbers = 5! = 120.
For finding total number of selections where 2 & 4 occur together, lets take 2&4 and bundle it to form one number X. So, total number of arrangements possible for 1,3,5 & x = 4! = 24.
But X is made up of 2 numbers 2 & 4, thus, number of ways X can be formed = 2! = 2.
So, total number of arrangements in which 2 & 4 are adjacent = 24 * 2 = 48.
Hence, total number of arrangements in which 2 & 4 are not adjacent = 120 - 48 = 72.
Cheers,
Dubes
Dubes
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kanha81
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This problem is quite similar to the problem of solving that neither of the 2 couples sit adjacent to each other of the given 5 people.
Total number of ways 5! = 120
x x x x x
1 2 3 4 5
Considering any of the above 2 numbers (except 2, 4) as a unit, we can arrange the 3 units in 3! = 6 ways
Now there are 2 such units of 2 that can be arranged in 2 * 2! = 4 ways
Hence, there are 6*4 = 24 ways to arrange the 5 numbers.
x x x x x
2 1 4 3 5
Similarly, there are another 24 ways to arrange 5 numbers following the above logic.
Hence, 120-48=72
Total number of ways 5! = 120
x x x x x
1 2 3 4 5
Considering any of the above 2 numbers (except 2, 4) as a unit, we can arrange the 3 units in 3! = 6 ways
Now there are 2 such units of 2 that can be arranged in 2 * 2! = 4 ways
Hence, there are 6*4 = 24 ways to arrange the 5 numbers.
x x x x x
2 1 4 3 5
Similarly, there are another 24 ways to arrange 5 numbers following the above logic.
Hence, 120-48=72
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