Prove
r*n∁r=n*(n-1)∁(r-1)
To prove the equality \(r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}\), we can use algebraic manipulations and the definition of the binomial coefficient.
The binomial coefficient \(\binom{n}{r}\) represents the number of ways to choose \(r\) objects out of a set of \(n\) objects, also denoted as "n choose r". It is defined as \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), where \(n!\) denotes the factorial of \(n\) (the product of all positive integers less than or equal to \(n\)).
Let's start with the left-hand side of the equation:
\(r \cdot \binom{n}{r}\)
Using the definition of the binomial coefficient, we can substitute \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\):
\(r \cdot \frac{n!}{r!(n-r)!}\)
Simplifying, we can cancel out the \(r!\) term:
\(\frac{r \cdot n!}{r \cdot (r-1)!(n-r)!}\)
The \(r\) in the numerator can also cancel with one of the \(r\) terms in the denominator:
\(\frac{n!}{(r-1)!(n-r)!}\)
Now, let's move on to the right-hand side of the equation:
\(n \cdot \binom{n-1}{r-1}\)
Again, using the definition of the binomial coefficient, we substitute \(\binom{n-1}{r-1} = \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}\):
\(n \cdot \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}\)
Simplifying, we can cancel out the \((r-1)!\) term:
\(\frac{n \cdot (n-1)!}{(r-1)!((n-1)-(r-1))!}\)
Notice that the numerator \(n \cdot (n-1)!\) is equivalent to \(n!\) since \(n!\) represents the factorial of \(n\):
\(\frac{n!}{(r-1)!((n-1)-(r-1))!}\)
Simplifying further, we have:
\(\frac{n!}{(r-1)! (n-r)!}\)
Comparing the right-hand side with the left-hand side, we see that they are equal:
\(\frac{n!}{(r-1)! (n-r)!} = \frac{n!}{(r-1)! (n-r)!}\)
Therefore, we have proven that \(r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}\).