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Question

Mathematics Question on Permutations

Find r if

(i) 5Pr = 26Pr-1

**(ii) **5Pr = 6Pr-1

Answer

(i) 5Pr = 26Pr-1

⇒5!(5−r)!=2×6!(6−r+1)!⇒\frac{5!}{\left(5-r\right)!}=2\times\frac{6!}{\left(6-r+1\right)!}

⇒5!(5−r)!=2×6!(7−r)!⇒\frac{5!}{\left(5-r\right)!}=2\times\frac{6!}{\left(7-r\right)!}

⇒5!(5−r)!=2×6×5!(7−r)(6−r)(5−r)!⇒\frac{5!}{\left(5-r\right)!}=\frac{2\times6\times5!}{\left(7-r\right)\left(6-r\right)\left(5-r\right)!}

⇒1=2×6(7−r)(6−r)⇒1=\frac{2\times6}{\left(7-r\right)\left(6-r\right)}
⇒(7−r)(6−r)=12⇒\left(7-r\right)\left(6-r\right)=12
⇒42−6r−7r+r2=12⇒42-6r-7r+r^2=12
⇒r2−13r+30=0⇒r^2-13r+30=0
⇒r2−3r−10r+30=0⇒r^2-3r-10r+30=0
⇒r(r−3)−10(r−3)=0⇒r\left(r-3\right)-10\left(r-3\right)=0
⇒(r−3)(r−10)=0⇒\left(r-3\right)\left(r-10\right)=0
⇒(r−3)=0  or(r−10)=0⇒\left(r-3\right)=0\space or \left(r-10\right)=0
⇒r=3⇒r=3 or r=10 r=10
nPr=n!(n−r)!^nP_r=\frac{n!}{\left(n-r\right)!}, where 0≤r≤n0≤r≤n
It is known that,
∴0≤r≤5∴0 ≤ r ≤ 5
Hence, r≠10∴r=3r \neq 10 ∴r = 3

(ii) 5Pr = 6Pr-1

⇒5!(5−r)!=6!(6−r+1)!⇒\frac{5!}{\left(5-r\right)!}=\frac{6!}{\left(6-r+1\right)!}

⇒5!(5−r)!=6×5!(7−r)!⇒\frac{5!}{\left(5-r\right)!}=6\times\frac{5!}{\left(7-r\right)!}

⇒1(5−r)!=6(7−r)(6−r)(5−r)!⇒\frac{1}{\left(5-r\right)!}=\frac{6}{\left(7-r\right)\left(6-r\right)\left(5-r\right)!}

⇒1=6(7−r)(6−r)⇒1=\frac{6}{\left(7-r\right)\left(6-r\right)}
⇒(7−r)(6−r)=6⇒\left(7-r\right)\left(6-r\right)=6
⇒42−7r−6r+r2−6=0⇒42-7r-6r+r^2-6=0
⇒r2−13r+36=0⇒r^2-13r+36=0
⇒r2−4r−9r+36=0⇒r^2-4r-9r+36=0
⇒r(r−4)−9(r−4)=0⇒r\left(r-4\right)-9\left(r-4\right)=0
⇒(r−4)(r−9)=0⇒\left(r-4\right)\left(r-9\right)=0
⇒(r−4)=0⇒\left(r-4\right)=0 or (r−9)=0\left(r-9\right)=0
⇒r=4⇒r=4 or r=9r=9

nPr=n!(n−r)!^nP_r=\frac{n!}{\left(n-r\right)!}, where0≤r≤n 0≤r≤n
It is known that,
0≤r≤50 ≤ r ≤ 5
Hence, r≠9r \neq 9
∴r=4∴ r = 4