To find the total number of arrangements where the even digits are not together, we can subtract the number of arrangements where the even digits are together from the total number of arrangements.
The total number of arrangements of the digits 1, 2, 3, 4, 5 is 5! = 5 x 4 x 3 x 2 x 1 = 120.
Now let's count the number of arrangements where the even digits (2 and 4) are together. We can treat 2 and 4 as a single entity, so we have a total of 4 entities: {24, 1, 3, 5}, {1, 24, 3, 5}, {1, 3, 24, 5}, {1, 3, 5, 24}. For each arrangement of the 4 entities, there are 2 ways to arrange the even digits within the {2, 4} entity. So the total number of arrangements where the even digits are together is 4 x 2 = 8.
Finally, we subtract the number of arrangements where the even digits are together from the total number of arrangements: 120 - 8 = <<120-8=112>>112.
Therefore, the total number of arrangements of the digits 1, 2, 3, 4, 5, if the even digits must not be together, is 112.
Which is the total number of arrangements of the digits 1, 2, 3, 4, 5, if the even digits must not be together?
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1 answer