Question

5. If "P, = 1680 and "C, = 70, find n and r.​

Answer 1

Answer:

npr=n!/(n-r)!

and ncr=n!/(n-r)!r!

given npr=1680 and ncr=70

n!/(n-r)!=1680 and n!/(n-r)!r!=70

1/r!×1680=70

r!=1680/70=24

r!=4!

r=4

np4=1680

n(n-1)(n-2)(n-3)=1680

(n^2-3n)(n^2-3n+2)=1680

let n^2-3n=t

t(t+2)=1680

t^2+2t-1680=0

t^2+42t-40t-1680=0

t(t+42)-40(t+42)=0

(t+42)(t-40)=0

t=40,-42

case1:If t=40 then n^2-3n-40=0

n^2-8n+5n-40=0

n(n-8)+5(n-8)=0

(n+5)(n-8)=0

n=8,-5

case2:if t=-42 then n^2-3n+42=0

sine D less than zero.The roots are imaginary.

Hence n=8