QUOTE(maximR @ Oct 16 2013, 02:30 PM)
QUOTE(maximR @ Oct 16 2013, 07:37 PM)
Sorry five choose four . Hmmm , after asking seniors , I think 1200 is the best answer . Do you have any more seniors whom you can double check with ? But I think I'm quite sure now , seeing your reasoning and manutd's .
QUOTE(maximR @ Oct 16 2013, 07:40 PM)
There should be only six letters in a code .
Permutations of 6-letter code in which the letter U & E are side by side:
P(6, U|E)(1) Use 2 letters :: permutations of U|E :: 2!
UE
EU
(2) Use the remaining 4 letters :: order of 4 out of 5-letter alphabet: S|B|J|C|T :: 5P4
SBJC
SBJT
SBCJ
SBCT
SBTJ
SBTC
...
(3) Ways of arranging 4 out of 5-slot code where the inseparable 2-letter U|E is treated as 1 fixed slot and the order of the slots doesn't matter :: 5C4
U|E _ _ _ _
_ U|E _ _ _
_ _ U|E _ _
_ _ _ U|E _
_ _ _ _ U|E
P(6, U|E) = (permutations of U|E) × (order of 4 out of 5-letter alphabet: S|B|J|C|T) × (ways of arranging 4 out of 5-slot code)
P(6, U|E) = 2! × 5P4 × 5C4 = 1200 different permutations
This post has been edited by Critical_Fallacy: Oct 17 2013, 08:46 AM