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 \(r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}\).