Factorials - Find the value of n if [2n! / 3! (2n - 3)!] : [n! / 2!(n - 2)!] = 44 : 3

Question:

Find the value of n if [2n! / 3! (2n - 3)!] : [n! / 2!(n - 2)!] = 44 : 3


Solution:

[2n! / 3! (2n - 3!)] : [n! / 2!(n - 2)!] = 44 : 3

2n! = 2n (2n - 1) (2n - 2) (2n - 3)!
3! = 3 x 2 = 6
n! = n (n - 1) (n - 2!)
2! = 2 x 1 = 2

∴ [2n (2n - 1) (2n - 2) (2n - 3)!] / [6 x (2n - 3)!] : [n (n - 1) (n - 2)!] / [2 x (n - 2)!] = 44 : 3

∴ [2n (2n - 1) (2n - 2) ̶(̶2̶n̶ ̶-̶ ̶3̶)̶!̶] / [6 x ̶(̶2̶n̶ ̶-̶ ̶3̶)̶!̶] : [n (n - 1) ̶(̶n̶ ̶-̶ ̶2̶)̶!̶] / [2 x ̶(̶n̶ ̶-̶ ̶2̶)̶!̶] = 44 : 3

∴ [2n (2n - 1) (2n - 2)] / [6] : [n (n - 1)] / [2] = 44 : 3

∴ [2n (2n - 1) 2 (n - 1)] / [6] : [n (n - 1)] / [2] = 44 : 3

∴ {[2n (2n - 1) 2 (n - 1)] / [6]} x {[2] / [n (n - 1)]} = 44 : 3

∴ {[2 ̶n̶ (2n - 1) 2 ̶(̶n̶ ̶-̶ ̶1̶)̶] / [6]} x {[2] / [ ̶n̶ ̶(̶n̶ ̶-̶ ̶1̶)̶]} = 44 : 3

∴ [2 (2n - 1) x 2 / 6] x 2 = 44 : 3

∴ [2 (2n - 1) / 3] x 2 = 44 : 3

∴ [4 (2n - 1) / 3] = 44 : 3

∴ [4 (2n - 1) / ̶3̶] = 44 : ̶3̶

∴ [4 (2n - 1)] = 44

∴ [8n - 4] = 44

∴ [8n] = 44 + 4

∴ [8n] = 48

∴ n = 6


Answer:

n = 6 if [2n! / 3! (2n - 3)!] : [n! / 2!(n - 2)!] = 44 : 3


Video Solution:



Find the value of n if [2n! / 3! (2n - 3)!] : [n! / 2!(n - 2)!] = 44 : A solved example on Factorials in Permutation and Combinations