Eclectic Enumeration

Logic Level 4

((a+b)+(c+d))(((a+b)+c)+d)((a+(b+c))+d)(a+(b+(c+d)))(a+((b+c)+d)) \large ((a+b)+(c+d)) \\ \large (( (a+b)+c) + d) \\ \large ( (a+(b+c)) + d ) \\ \large ( a+(b+(c+d))) \\ \large (a + ((b+c)+d) )

Suppose I'm given 4 numbers and I'm given a task to put in parentheses in it in such a way that I'm only adding two numbers at a time, then there will a total of 5 ways to do it as described above.

What would the answer be if I'm given 8 numbers instead?

Details and Assumptions:

  • As an explicit example, if the numbers are a1,a2,,a8a_1, a_2,\ldots, a_8, then (((a1+a2)+(a3+a4))+((a5+a6)+(a7+a8))) (((a_1+a_2)+(a_3+a_4))+((a_5+a_6)+(a_7+a_8))) is allowed because we are adding 2 numbers at a time. But (((a1+a2+a3)+(a4+a5))+((a6+a7)+a8)) (( (a_1 + a_2 + a_3) + (a_4 + a_5)) + ((a_6 + a_7) + a_8)) is not allowed because we're adding more than 2 numbers at a time.

  • You must keep the numbers a1,a2,,a8a_1,a_2,\ldots,a_8 in that order (from left to right).


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