When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated expression, there may be several brackets nested amongst each other, such as in the expression (x+1)∗((x−2)+3(x−4)×(x^2+7×(3x+4))). If we removed all the symbols other than the brackers from the expression, we would be left with the arrangement ()(()()(())). For any arrangement of brackets, it could have come from a valid mathematical expression if and only if for every place in the sequence, the number of open brackets before that place is at least as large as the number of closed brackets. If 34 open brackets and 34 closed brackets are randomly arranged, the probability that the resulting arrangement could have come from a valid mathematical expression can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?

Question

When writing a math expression, any time there is an open bracket "(", it is eventually followed by a closed bracket ")". When we have a complicated expression, there may be several brackets nested amongst each other, such as in the expression (x+1)∗((x−2)+3(x−4)×(x^2+7×(3x+4))). If we removed all the symbols other than the brackers from the expression, we would be left with the arrangement ()(()()(())). For any arrangement of brackets, it could have come from a valid mathematical expression if and only if for every place in the sequence, the number of open brackets before that place is at least as large as the number of closed brackets. If 34 open brackets and 34 closed brackets are randomly arranged, the probability that the resulting arrangement could have come from a valid mathematical expression can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?

Answer 1

no need to cheat on brilliant, don't answer

Answer 2

To determine the probability that a randomly arranged sequence of 34 open brackets and 34 closed brackets could have come from a valid mathematical expression, we need to apply the condition mentioned in the question: for every place in the sequence, the number of open brackets before that place should be at least as large as the number of closed brackets.

To solve this problem, we can use a principle known as Catalan numbers. Catalan numbers represent the number of valid bracket sequences of a specific length.

The formula to calculate the nth Catalan number is given by:
C(n) = (2n)! / ((n+1)! * n!)

In our case, we have 34 open brackets and 34 closed brackets, so we need to calculate C(34).

Using the formula, we have:
C(34) = (2 * 34)! / ((34 + 1)! * 34!)

Simplifying this expression, we get:
C(34) = 68! / (35! * 34!)

To find the probability, we need to divide the number of valid bracket sequences (C(34)) by the total number of possible arrangements (which is the number of ways to arrange 34 open brackets and 34 closed brackets).

The total number of arrangements is given by:
Total arrangements = (34 + 34)! = 68!

Therefore, the probability is:
Probability = C(34) / Total arrangements = C(34) / (68!)
= 68! / (35! * 34!) / 68!
= 1 / (35!)

Since we need to express the probability as a fraction, we have:
Probability = 1 / (35!) = 1/10333147966386144929666651337523200000000

The value of a+b is 1 + 10333147966386144929666651337523200000000 = 10333147966386144929666651337523200000001.