If { }^{n} P_{r}={ }^{n} P_{r+1} and { }^{n} | vedclass

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(D) Given ${ }^{n} P_{r}={ }^{n} P_{r+1}$,we have:$\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$Since $n! 
eq 0$,we can divide both sides by $n!$:$\frac{1

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$2$

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(D) Given ${ }^{n} P_{r}={ }^{n} P_{r+1}$,we have:$\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$Since $n! 
eq 0$,we can divide both sides by $n!$:$\frac{1}{(n-r)(n-r-1)!} = \frac{1}{(n-r-1)!}$$n-r = 1 \Rightarrow n = r+1$  $(1)$Given ${ }^{n} C_{r}={ }^{n} C_{r-1}$,we have:$\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}$$\frac{1}{r(r-1)!(n-r)!} = \frac{1}{(r-1)!(n-r+1)(n-r)!}$$\frac{1}{r} = \frac{1}{n-r+1}$$n-r+1 = r \Rightarrow n+1 = 2r$  $(2)$Substitute $n = r+1$ from $(1)$ into $(2)$:$(r+1)+1 = 2r$$r+2 = 2r$$r = 2$

If ${ }^{n} P_{r}={ }^{n} P_{r+1}$ and ${ }^{n} C_{r}={ }^{n} C_{r-1}$,then the value of $r$ is equal to:

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