How many permutations of the numbers 1 through 8 are there in which none of the terms 12, 34, 56, 78 appears?

Step-by-step explanation:

Let us assume that Aij is the set of all orderings in which the pattern ij appears.

Then the total number of orderings of numbers 1, 2 . . . , 8 is a total of 8!.

we are then asked to look for

8! − |A12 ∪ A34 ∪ A56 ∪ A78| (using the Addition Principle).

Again, if we use the Inclusion-Exclusion Principle, we have

|A12 ∪ A34 ∪ A56 ∪ A78| = |A12| + |A34| + |A56| + |A78| − |A12 ∩ A34| − |A12 ∩ A56| − |A12 ∩ A78| − |A34 ∩ A56| − |A34 ∩ A78| − |A56 ∩ A78| + |A12 ∩ A34 ∩ A56| + |A12 ∩ A34 ∩ A78| + |A34 ∩ A56 ∩ A78| − |A12 ∩ A34 ∩ A56 ∩ A78|

To find the side of A12, we treat 12 as a single number, as a result, we need to find the number of ways we can order 7 numbers 12, 3, 4, 5, 6, 7, 8. This is 7!, hence |A12| = 7!.

Using the same method, we find that |A34| = |A56| = |A78| = 7!.

Now, to find |A12 ∩ A34|, we need to find the number of orderings that fix 12 and 34.

To do that, we treat 12 as one number, 34 as a different one, then |A12 ∩ A34| is equal to the number of orderings of 6 numbers, i.e., |A12 ∩ A34| = 6!.

In the same way, we realize that |A12 ∩ A56| = |A12 ∩ A78| = |A34 ∩ A56| = |A34 ∩ A78| = |A56 ∩ A78| = 6!.

This means that |A12 ∩ A34 ∩ A56| = |A12 ∩ A34 ∩ A78| = |A34 ∩ A56 ∩ A78| = 5! and |A12 ∩ A34 ∩ A56 ∩ A78| = 4!.

Therefore,

|A12 ∪ A34 ∪ A56 ∪ A78| = 4(7!) − 6(6!) + 4(5!) − 4!

and

8! − |A12 ∪ A34 ∪ A56 ∪ A78| = 8! − 4(7!) + 6(6!) − 4(5!) + 4!

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10 you would take 50/ .2

step-by-step explanation: