If n c r-1 = 36, n c r =84 and n c r+1 = 126, find r c 2
Answers
Step-by-step explanation:
NSWER
\cfrac{^nC_{r}}{^nC_{r-1}}=\cfrac{84}{36}
n
C
r−1
n
C
r
=
36
84
\Rightarrow\cfrac{\cfrac{n!}{r!(n-r)!}}{\cfrac{n!}{(r-1)!(n-r+1)!}}=\cfrac{84}{36}⇒
(r−1)!(n−r+1)!
n!
r!(n−r)!
n!
=
36
84
\Rightarrow\cfrac{n-r+1}{r}=\cfrac{7}{3}⇒
r
n−r+1
=
3
7
\Rightarrow 3n-3r+3=7r⇒3n−3r+3=7r
\Rightarrow10r=3n+3⇒10r=3n+3---------------1
\cfrac{^nC_{r+1}}{^nC_{r}}=\cfrac{126}{84}
n
C
r
n
C
r+1
=
84
126
\Rightarrow\cfrac{\cfrac{n!}{(r+1)!(n-r-1)!}}{\cfrac{n!}{(r)!(n-r)!}}=\cfrac{3}{2}⇒
(r)!(n−r)!
n!
(r+1)!(n−r−1)!
n!
=
2
3
\Rightarrow \cfrac{n-r}{r+1}=\cfrac{3}{2}⇒
r+1
n−r
=
2
3
\Rightarrow 2n-2r=3r+3⇒2n−2r=3r+3
\Rightarrow 5r=2n-3⇒5r=2n−3-------------------2
Dividing Equation 1 by Equation 2,
\cfrac{10r}{5r}=\cfrac{3n+3}{2n-3}
5r
10r
=
2n−3
3n+3
\Rightarrow2(2n-3)=3n+3⇒2(2n−3)=3n+3
\Rightarrow4n-6=3n+3⇒4n−6=3n+3
\Rightarrow n=6+3⇒n=6+3
\Rightarrow n=9⇒n=9
10r=3n+310r=3n+3
\Rightarrow 3(9)+3=30⇒3(9)+3=30
\Rightarrow 10r=30⇒10r=30
\Rightarrow r=3⇒r=3
\therefore\, n=9,r=3∴n=9,r=3