If \(n\) and \(r\) are non-negative integers and \(n ≥ r\), then \(p(n + 1,…
2013
If \(n\) and \(r\) are non-negative integers and \(n ≥ r\), then \(p(n + 1, r)\) equals to
- A.
\(\frac {p(n, r) (n + 1)} {(n + 1 -r)}\) - B.
\(\frac {p(n, r) (n + 1)} {(n - 1 + r)}\) - C.
\(\frac {p(n, r) (n - 1)} {(n + 1 -r)}\) - D.
\(\frac {p(n, r) (n + 1)} {(n + 1 + r)}\)
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Correct answer: A
Answer: p(n+1, r) = (n+1)!/(n+1-r)! = (n+1)/(n+1-r) · p(n, r).
Start from the definitions: p(n, r) = n!/(n-r)! and p(n+1, r) = (n+1)!/(n+1-r)!.
Use factorial identities: (n+1)! = (n+1)·n! and (n+1-r)! = (n+1-r)·(n-r)!.
Substitute these into p(n+1, r): p(n+1, r) = ((n+1)n!)/((n+1-r)(n-r)!) = (n+1)/(n+1-r) · (n!/(n-r)!) = (n+1)/(n+1-r) · p(n, r).
Thus the expression equals p(n, r)·(n+1)/(n+1-r).