If a = 9/5, and 2, 3, 7, only 4 arrangements work.
1) (7-2)x(3+a)
2) (3+a)x(7-2)
3) (a+3)x(7-2)
4) (7-2)x(a+3)
Someone may suggest that this represents only one (1) solution.
Programming is a wonderful thing. Just to get you started:
© Computes all possibilities of 4 numbers combined with the
© 4 basic arithmetic functions, with every possible association.
© 4 × 4 × 4 × 4 × 4 × 4 × 4 = 16384 initial possibilities
© 4 × 4 × 3 × 4 × 2 × 4 × 1 = 1536 actual possibilities
© 1) A×(B×(C×D)) - '×' as used here, represents any of ('×','÷','-','+')
© 2) (A×B)×(C×D) - A sixth variety is identical to this one.
© 3) A×((B×C)×D)
© 4) (A×(B×C))×D
© 5) ((A×B)×C)×D
Don't forget to check for redundancy and for illegal operations (such as division by zero).