If \[{}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52\], find r.

Hint:The expression is that of Permutation, which represents ordered matters. For number of permutation of n things taken r at a time = \[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]. Simplify the given expression with this formula and find the value of r.

Complete step-by-step answer:
Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. Permutation is also the linear order of an ordered set. Thus the number of permutation (ordered matters) of n things taken r at a time is given as,
\[{}^{n}{{P}_{r}}=P\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}\]
Now, we have been given that,
\[{}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52…...(1)\]
Let us simplify it as per the formula of Permutation.
\[{}^{22}{{P}_{r+1}}=\dfrac{22!}{\left( 22-r-1 \right)!}=\dfrac{22!}{\left( 21-r \right)!}\]
Similarly, \[{}^{20}{{P}_{r+2}}=\dfrac{20!}{\left( 20-r-2 \right)!}=\dfrac{20!}{\left( 18-r \right)!}\]
Thus substitute back the simplified expression of \[{}^{22}{{P}_{r+1}}\] and \[{}^{20}{{P}_{r+2}}\] in (1)
\[{}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52\]
\[\Rightarrow \dfrac{22!}{\left( 21-r \right)!}:\dfrac{20!}{\left( 18-r \right)!}=11:52\]
Thus we can write the above as,
\[\dfrac{\dfrac{22!}{\left( 21-r \right)!}}{\dfrac{20!}{\left( 18-r \right)!}}=\dfrac{11}{52}\]
\[\Rightarrow \dfrac{22!}{\left( 21-r \right)!}\times \dfrac{\left( 18-r \right)!}{20!}=\dfrac{11}{52}\]
We can write \[22!=22\times 21\times 20!\]
Similarly we can write, \[\left( 21-r \right)!=\left( 21-r \right)\left( 20-r \right)\left( 19-r \right)\left( 18-r \right)!\]
Thus we can write (1) as,
\[\dfrac{22\times 21\times 20!}{\left( 21-r \right)\left( 20-r \right)\left( 19-r \right)\left( 18-r \right)!}\times \dfrac{\left( 18-r \right)!}{20!}=\dfrac{11}{52}\]
Now cancel out the like terms and cross multiply it.
\[\begin{align}
  & \dfrac{22\times 21}{\left( 21-r \right)\left( 20-r \right)\left( 19-r \right)}=\dfrac{11}{52} \\
 & \Rightarrow 22\times 21\times 52=11\left( 21-r \right)\left( 20-r \right)\left( 19-r \right) \\
 & \Rightarrow \dfrac{22\times 21\times 52}{11}=\left( 21-r \right)\left( 20-r \right)\left( 19-r \right) \\
 & \therefore \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=2\times 21\times 52 \\
 & \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=2\times \left( 3\times 7 \right)\times \left( 4\times 13 \right) \\
 & \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=\left( 2\times 7 \right)\times 13\times \left( 3\times 4 \right) \\
 & \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=14\times 13\times 12 \\
\end{align}\]
Now, 21 – 7 = 14
Similarly, 20 – 7 = 13 and 19 – 7 = 12
\[\begin{align}
  & \therefore \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=14\times 13\times 12 \\
 & \Rightarrow \left( 21-r \right)\left( 20-r \right)\left( 19-r \right)=\left( 21-7 \right)\left( 20-7 \right)\left( 19-7 \right) \\
\end{align}\]
Thus we can say that, r = 7.
Hence we get the value of r = 7.

Note: Don’t confuse the formula of permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n things taken r at a time:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].We can also verify the answer by substituting value r=7 in the expression \[{}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52\] and check whether L.H.S=R.H.S or not.