prove (n choose k) = (n choose n-k)
it's a definition
nCk = n! / [k! * (n-k)!]
nC(n-k) = n! / {(n-k)! * [n - (n-k)]!}
n - (n-k) = k
To prove that (n choose k) = (n choose n-k), we can use the formula for the binomial coefficient, also known as "n choose k":
(n choose k) = n!/[(n-k)! * k!]
where "n!" denotes the factorial of n, which is the product of all positive integers up to n.
Similarly, we can write (n choose n-k) as:
(n choose n-k) = n!/[(n-(n-k))! * (n-k)!] = n!/[(k)! * (n-k)!]
So our goal is to show that n!/[(n-k)! * k!] = n!/[(k)! * (n-k)!].
To prove this equation, we can simplify both sides and show they are equal:
n!/[(n-k)! * k!] = n!/[(k)! * (n-k)!]
Cancelling out the common factors from both sides, we get:
(n-k)! * k! = (k)! * (n-k)!
Now, since (n-k)! and (k)! are factorials, they are both products of consecutive positive integers. Therefore, they are equal to each other:
(n-k)! = (k)!
Substituting this equality back into our equation, we have:
(n-k)! * k! = (k)! * (n-k)!
(k)! * (n-k)! = (k)! * (n-k)!
Both sides of the equation are identical, so we have proven that:
(n choose k) = (n choose n-k)
Thus, we have proved our statement.