Question 1208345
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In how many ways can the digits of the number 30348877 be arranged such that 
no two even digits are adjacent?
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<pre>
We have four even digits  0, 4, 8, 8  and 4 odd digits  3, 3, 7, 7.


Two possible placements are  EOEOEOEO  or  OEOEOEOE,  where E is the placeholder 
for an even digit and O is the placeholder for an odd digit.


For odd digits with repetitions, as in the given number, there are

    {{{4!/(2!*2!)}}} = {{{24/(2*2)}}} = 6 different distinguishable permutations are possible 

inside the group of odd numbers.


For even digits with repetitions, as in the given number, there are

    {{{4!/(2!)}}} = {{{24/2}}} = 12 different distinguishable permutations are possible 

inside the group of even numbers.


Of these 12 arrangements, the number of those that start with 0 is 3;  so these 3 should be subtracted from 12.
Doing this way, we find that the number of arrangements of four even numbers in this problems, 
that do not start from 0 is 12-3 = 9.



Thus we have 6*12 = 72 arrangements of the type OEOEOEOE and 9*6 = 54 arrangements of the type EOEOEOEO.


The total is the sum 72 + 54 = 126.



Hence, in total, there are  126 different distinguishable arrangements as required in the problem.   <U>ANSWER</U>
</pre>

Solved.