Solution :
We have, P(n, r) = \(n!\over (n – r)!\)
Putting r = n,
\(\implies\) P(n, n) = \(n!\over 0!\)
\(\implies\) n! = \(n!\over 0!\) [ \(\because\) P(n, n) = n! ]
\(\implies\) 0! = \(n!\over n!\) = 1
Hence, Proved.
We have, P(n, r) = \(n!\over (n – r)!\)
Putting r = n,
\(\implies\) P(n, n) = \(n!\over 0!\)
\(\implies\) n! = \(n!\over 0!\) [ \(\because\) P(n, n) = n! ]
\(\implies\) 0! = \(n!\over n!\) = 1
Hence, Proved.