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

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

Explanation

Solution

Hint:The expression is that of Permutation, which represents ordered matters. For number of permutation of n things taken r at a time = nPr=n!(n−r)!{}^{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,
nPr=P(n,r)n!(n−r)!{}^{n}{{P}_{r}}=P\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}
Now, we have been given that,
22Pr−1:20Pr+2=11:52…...(1){}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52…...(1)
Let us simplify it as per the formula of Permutation.
22Pr+1=22!(22−r−1)!=22!(21−r)!{}^{22}{{P}_{r+1}}=\dfrac{22!}{\left( 22-r-1 \right)!}=\dfrac{22!}{\left( 21-r \right)!}
Similarly, 20Pr+2=20!(20−r−2)!=20!(18−r)!{}^{20}{{P}_{r+2}}=\dfrac{20!}{\left( 20-r-2 \right)!}=\dfrac{20!}{\left( 18-r \right)!}
Thus substitute back the simplified expression of 22Pr+1{}^{22}{{P}_{r+1}} and 20Pr+2{}^{20}{{P}_{r+2}} in (1)
22Pr−1:20Pr+2=11:52{}^{22}{{P}_{r-1}}:{}^{20}{{P}_{r+2}}=11:52
⇒22!(21−r)!:20!(18−r)!=11:52\Rightarrow \dfrac{22!}{\left( 21-r \right)!}:\dfrac{20!}{\left( 18-r \right)!}=11:52
Thus we can write the above as,
22!(21−r)!20!(18−r)!=1152\dfrac{\dfrac{22!}{\left( 21-r \right)!}}{\dfrac{20!}{\left( 18-r \right)!}}=\dfrac{11}{52}
⇒22!(21−r)!×(18−r)!20!=1152\Rightarrow \dfrac{22!}{\left( 21-r \right)!}\times \dfrac{\left( 18-r \right)!}{20!}=\dfrac{11}{52}
We can write 22!=22×21×20!22!=22\times 21\times 20!
Similarly we can write, (21−r)!=(21−r)(20−r)(19−r)(18−r)!\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,
22×21×20!(21−r)(20−r)(19−r)(18−r)!×(18−r)!20!=1152\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.

& \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{aligned}$$ Now, 21 – 7 = 14 Similarly, 20 – 7 = 13 and 19 – 7 = 12 $$\begin{aligned} & \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{aligned}$$ Thus we can say that, r = 7. Hence we get the value of r = 7. Note: Don’t confuse the formula of permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n things taken r at a time: $${}^{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.