Question 1208345
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There are 8 digits of which 4 are even and 4 are odd. If no two even digits are adjacent, then the digits must alternate between even and odd.<br>
I will use a different path to the answer than the other tutor.  Maybe an easier path, maybe not; just different.<br>
We can choose any of the 8 digits to be first.  After that, since the even and odd digits must alternate, the numbers of choices for the subsequent digits are 4, 3, 3, 2, 2, 1, and 1.   So the number of sequences in which we can choose the digits is 8*4*3*3*2*2*1*1.<br>
But there are 3 pairs of identical digits.  Because in each of those pairs the identical digits can be in either of 2 orders, this number of arrangements is too big by a factor of 2*2*2.  So the number of ways of arranging the digits of the number is<br>
{{{(8*4*3*3*2*2*1*1)/(2*2*2)=144}}}<br>
ANSWER: 144<br>
Note the other tutor doesn't allow 0 as a first digit.  But the problem doesn't specifically preclude that; there are many applications where a "number" can have a leading 0.<br>
If we don't allow 0 as the first digit, then the calculation shown above has 7 instead of 8 choices for the first digit, which leads to the answer of 126 obtained by the other tutor by her method.<br>