Question

Take any 5 Numbers from the brackets(1 3 5 7 9 11 13 15) and solve it to get a sum of 30.numbers can be repeated

Answer 1

Gaurav Agrawal answered very intelligently and using his imagination.         -- really for this question is asked so many times.. all credit goes to him. 
                           3! + 3! + 3! + 3! + 3!      = 30 

You can answer this question in many ways using the factorial function.  The following are my answers, taking the tip from his answer.

   3! + 1 + 3 + 9 + 11          = 30
       1 + 5 + 7 + 3! + 11               = 30
             3! + 3! + 3! + 11 + 1             = 30
                 3! + 3! + 11 + 5 + 1               = 30
            11+ 1 + 1 + 3! + 11              = 30
       15 + 3 + 3 + 3 + 3!              = 30
     13 + 1 + 1 + 9 + 3!            = 30
         13 + 1 + 3 + 7 + 3!            = 30
               13 + 5 + 3 + 3 + 3!            = 30

If  one can use /, * , exponent and other operators, then you can get more answers too.  Question does not say anything about the operators.

  5 * 3 + 13 + 1 + 1 = 30
     13 + 15 + 3! - 3 - 1 =    30
          15! / 13! / 7 + 1 - 1  =  30
               5!  / 3!  + 1 * 3 + 7 = 30
                      7 * 5  - 3 - 1 - 1  =  30
                   9 * 7 / 3 + 9 * 1 =  30
          3³ + 1 + 11 - 13 * 1  = 30
     5³ / 15 * 3 + 5 * 1 = 30  
 3 * 15 - 11 - 3 - 1 = 30

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You need one operator for which the result is an even number with an input being an odd number.  Factorial is one.  Probably there are other operators too.

Answer 2

3!+3!+3!+3!+3!=30

Note that 3!=3*2*1=6