Question 629274
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Different positive four-digit integers are to be formed by using each of the digits 1,2,3,4 just once 
in each integer. How many different such integers can be formed if the digits 3 and 4 must NEVER be next to each other?
A. 4
B. 8
C. 12
D. 16
E. 24
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        The idea is to calculate the number of all possible permutations of the four given digits

        and then subtract all unfavorable arrangements,  where  3  and  4  are next to each other.



<pre>
The number of all possible permutations of the four given digits is 4! = 1*2*3*4 = 24.


The unfavorable arrangements are of two kinds.


    One kind of unfavorable arrangements is when adjacent 3 and 4 go in this order, '34'.

    Then we consider this block as one entity X, and we have 3 (three) objects to permutate:
          the digits 1, 2 and the block X.  
          It gives 3! = 1*2*3 = 6 four-digit numbers with the block '34'.


    The other kind of such arrangements is when adjacent 3 and 4 go in this order, '43'.

    Then we consider this block as one entity Y, and we have 3 objects to permutate:
          the digits 1, 2 and the block Y.  
          It gives 3! = 1*2*3 = 6 four-digit numbers with the block '43'.


Thus the number of favorable arrangements is  4! - 2*3! = 24 - 2*6 = 24 - 12 = 12.


<U>ANSWER</U>.  The number of favorable arrangements is 12.
</pre>

Solved.