Solution:

It is given that : P (11, r) = P (12, r – 1)

By making use of the formula,

$ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $

$ P\text{ }\left( 11,\text{ }r \right)\text{ }=\text{ }11!/\left( 11-r \right)! $

$ P\text{ }\left( 12,\text{ }r-1 \right)\text{ }=\text{ }12!/\left( 12-\left( r-1 \right) \right)! $

$ =12!/\left( 12-r+1 \right)! $

$ =12!/\left( 13-r \right)! $

So, according to the question,

P (11, r) = P (12, r – 1)

Upon putting the obtained values in the above expression we get,

11!/(11 – r)! = 12!/(13 – r)!

Upon evaluating further, we get:

$ \left( 13-r \right)!\text{ }/\text{ }\left( 11-r \right)!\text{ }=\text{ }12!/11! $

$ \left[ \left( 13-r \right)\text{ }\left( 13-r-1 \right)\text{ }\left( 13-r-2 \right)! \right]\text{ }/\text{ }\left( 11\text{ }\text{ }r \right)!\text{ }=\text{ }\left( 12\times 11! \right)/11! $

$ \left[ \left( 13-r \right)\text{ }\left( 12-r \right)\text{ }\left( 11-r \right)! \right]\text{ }/\text{ }\left( 11-r \right)!\text{ }=\text{ }12 $

$ \left( 13-r \right)\text{ }\left( 12-r \right)\text{ }=\text{ }12 $

$ 156-12r-13r\text{ }+\text{ }{{r}^{2}}~=\text{ }12 $

$ 156-12-25r\text{ }+\text{ }{{r}^{2}}~=\text{ }0 $

$ {{r}^{2}}-25r\text{ }+\text{ }144\text{ }=\text{ }0 $

$ {{r}^{2}}-16r-9r\text{ }+\text{ }144\text{ }=\text{ }0 $

$ r\left( r-16 \right)-9\left( r-16 \right)=0 $

$ \left( r-9 \right)\text{ }\left( r-16 \right)\text{ }=\text{ }0 $

$ r\text{ }=\text{ }9\text{ }or\text{ }16 $

For, P (n, r): r ≤ n

∴ r = 9 [for, P (11, r)]