Question

An ice cream parlor has ice creams in eight different varieties. Number of ways of choosing 3 ice creams taking two ice creams of the same variety is:
$
  {\text{A}}{\text{. 56}} \\
  {\text{B}}{\text{. 64}} \\
  {\text{C}}{\text{. 100}} \\
  {\text{D}}{\text{. None of the above}} \\
$

Answer 1

Hint: First we find the possibility that the three ice creams chosen are of the same variety. Then we find the possibility that the two ice creams are of the same variety and the third is of another variety and we add the two cases.

Complete step-by-step answer:
Given Data – Choose 3 ice creams where at least two ice creams are of the same variety.
(At least = minimum)
So, there are two possibilities.
The first possibility is that the three ice creams chosen are of the same variety.
Number of ways of choosing 3 ice creams which are of the same variety ${}^{\text{n}}{{\text{P}}_{\text{r}}}$, here n = 8 and r = 1
Hence, ${}^{\text{n}}{{\text{P}}_{\text{r}}}$= $\dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}} = \dfrac{{8!}}{{\left( {8 - 1} \right)!}} = \dfrac{{8!}}{{7!}} = 8 $ ways -- (1)

The second possibility is that the two ice creams are of the same variety and the third is of another variety.
Number of ways choosing 2 ice creams of same variety and third ice cream of another variety is
${}^8{{\text{P}}_1}.{}^7{{\text{P}}_1} = {\text{ }}\dfrac{{8!}}{{\left( {8 - 1} \right)!}}.\dfrac{{7!}}{{\left( {7 - 1} \right)!}} = 8 \times 7 = 56 $ ways -- (2)

Hence, number of ways of choosing 3 ice creams taking at least two ice creams of the same variety = (1) + (2) = 8 + 56 = 64 ways.
Hence Option B is the correct answer.

Note – In order to solve this type of question the key is to carefully count the number of possibilities it can be done, two in this case. It is vital to identify that the selection is a permutation, i.e. the order is not important, using the permutation formula for each case individually we find their possibilities and add them for the answer.