Question 1152189
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The leftmost position digit can not be 0 --- hence, it is 1, since there is no other possibilities.


The rightmost position (the "ones" position) must be 0 to provide an even number.


Hence, you have 8 remaining positions to fill with six ones and two zeros by an arbitrary way.


You can place two digits of zero among 8 positions by  {{{C[8]^2}}} = {{{(8*7)/2}}} different ways = 4*7 = 28 different ways,


It gives you 28 different 10-digit numbers  under the question.    <U>ANSWER</U>
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