Question

An ordinary cubical dice having six faces marked with alphabets $A,B,C,D,E$ and $F$ is thrown $n$ times and the list of $n$ alphabets showing up are noted. Find the total number of ways in which among the alphabets $A,B,C,D,E$ and $F$ only three of them appear in the list.

Answer 1

Out of six faces, three can be selected in ${}^6{C}_3$ ways.

Consider one such selection, say $ABC$. Each of the ${}^{\prime }{n}^{\prime }$ places can be filled in three ways.

So total number of ways is $3^n$.

But this includes those ways also, which contain exactly one alphabet or exactly two alphabets which are to be subtracted.

Now, number of ways which contain only one letter is $3$ and number of ways containing exactly two alphabets is ${}^3{C}_2(2^n-2)$.

Hence, the number of ways is $3^n{-}^3{C}_2(2^n-2)-3$.

So, required number of ways is ${}^6{C}_3\left[3^n{-}^3{C}_2\right(2^n-2)-3]$.