2nC3:nC2=44:3? Permutation And book answer is 6
1 Answer
May 27, 2018
#n=6#
Explanation:
We seek
Using the definition of
#""_nC^r = (n!)/(r!(n-r)!) #
We can write the given ratio expression as :
# ((2n)!)/(3!(2n-3)!) \ : \ (n!)/(2!(n-2)!) = 44 \ : \ 3#
Or, Equivalently as a fraction:
# ( ((2n)!)/(3!(2n-3)!) ) / ( (n!)/(2!(n-2)!) )= 44 / 3#
So that:
# ((2n)!)/(3!(2n-3)!) * (2!(n-2)!)/(n!) = 44 / 3#
And if we expand:
# ((2n)(2n-1)(2n-2)(2n-3)....1)/(3!(2n-3)!) * (2!(n-2)!)/(n(n-1)(n-2)...1) = 44 / 3#
Cancelling terms, we get:
# ((2n)(2n-1)(2n-2))/(3!) * (2!)/(n(n-1)) = 44 / 3#
And Simplifying:
# (2(n)(2n-1)2(n-1))/(6) * (2)/(n(n-1)) = 44 / 3#
# :. (4(2n-1))/(3) = 44 / 3#
# :. 2n-1 = 11 #
# :. n=6 #
And we can quickly validate the solutions:
#""_(12)C^3 \ : \ _6C^2 = 220:15 = 44:3#
