2nC3:nC2=44:3? Permutation And book answer is 6

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Answer 1 · 2018-05-27T11:33:50

#n=6#

We seek #n in NN# st #""_(2n)C^3:_nC^2=44:3#

Using the definition of #""_nC^r#

#""_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#