Question

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

Answer 1

Hint:
Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.
The formula for permutations is: $\mathrm{{}^{n}P_{r}}= \dfrac{n!}{(n-r)!}$
The formula for combinations is: $\mathrm{{}^{n}C_{r}}= \dfrac{n!}{r!(n-r)!}$
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, we can select the items in any order. Combinations can be confused with permutations. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.

Complete step by step solution:
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, we can select the items in any order. Combinations can be confused with permutations. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.
Out of six faces, three can be selected in ${ }^{6} \mathrm{C}_{3}$ ways.
Consider one such selection, say $\mathrm{ABC.}$ Each of the ${ }^{\prime} \mathrm{n}^{\prime}$ places can
be filled in three ways.
So the total number of ways is $3^{\mathrm{n}}$.
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 ${ }^{3} \mathrm{C}_{2}\left(2^{\mathrm{n}}-2\right)$.
Hence, the number of ways is $3^{\mathrm{n}}-{ }^{3} \mathrm{C}_{2}\left(2^{\mathrm{n}}-2\right)-3$

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

Note:
One could say that a permutation is an ordered combination. The number of permutations of $\mathrm{n}$ objects taken $\mathrm{r}$ at a time is determined by the following formula:
$\mathrm{P}(\mathrm{n}, \mathrm{r})=\dfrac{n!}{(n-r)!}$. $\mathrm{n} !$ is read $\mathrm{n}$ factorial and means all numbers from 1 to $\mathrm{n}$ multiplied. Combinations are a way to calculate the total outcomes of an event where order of
the outcomes do not matter.
To calculate combinations, we will use the formula $\mathrm{{}^{n}C_{r}}= \dfrac{n!}{r!(n-r)!}$, where $\mathrm{n}$ represents the total number of items, and $\mathrm{r}$ represents the number of items being chosen at a time. Thus, $\mathrm{{}^{n}P_{r}}, \mathrm{r}$ The number of possibilities for choosing an ordered set of $\mathrm{r}$ objects $(\mathrm{a}$ permutation) from a total of n objects.