TIFR CSE 2012 | Part A | Question: 9

Question

The probability of throwing six perfect dices and getting six different faces is

• A. $1 -\dfrac{ 6!} { 6^{6}}$
• B. $\dfrac{6! }{ 6^{6}}$
• C. $6^{-6}$
• D. $1 - 6^{-6}$
• E. None of the above

Answer 1

$B)$ for all the $6$ different faces $1,2,3,4,5,6$

the probability is $\dfrac{1}{6^{6}} $ and then for $6!$ different permutations $:\dfrac{6!}{6^{6}}.$

Comments

oh yes , it may be case when only one face will different ... thanx ... :)

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For more explanation , refer to https://www.youtube.com/watch?v=T59f2xTHck8

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Yes it won't because, what you are saying is 1 - p(all getting same) = this may give you arrangements like (1,2,3,4,5,1) and many more such arrangements.. so it's better to arrange all in such a way that no two are having same number. i.e if we have two numbers( 1,2 ) and 2 places then we can only 2! arrangments same works with 6 numbers in 6 places- 6! ways..

Answer 2

we have 6 dices 
Total possible combination 6⁶
Total arrangement of different sides ⁶P₆ 
Therefore Probability = Total arrangement of different sides / Total possible combination 
                               =  6! / 6⁶ Option B 
 

Comments

Sir can you please explain the line "Total arrangement of different sides" is 6P6 ? I am not getting the solution.

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Sir, I got your explanation. Thanks sir