Question

A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in this stock. There are 5 places in a row in his showcase. What is the number of different ways of displaying the three varieties of perfumes in the showcase?
A. 6
B. 50
C. 150
D. None of these

Answer 1

Hint: To solve the above question first we will make two different cases first case will have to be 3 of 1 kind of perfume and 2 different kinds of perfume, and the second case will be there will be 2 pairs of 1 kind of perfume and 1 different kinds of perfume, and finally by adding the result of both cases we will find the required result.

Formula used: We will use Combination formula which is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], and factorial formula i.e., \[n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times .........1\]

Complete Step-by Step Solution:
Given that a shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in this stock. There are 5 places in a row in his showcase.
So, from the given data there are 5 places in the shopkeeper showcase and has three varieties of perfumes. We can conclude that there would be two cases.
Case I: There will be 3 of 1 kind of perfume and 2 different kinds of perfume,
\[\therefore \]Number of arrangements of 3 of 1 kind of perfume and 2 different kind of perfume will be
\[ \Rightarrow {}^3{C_1} \times {}^2{C_2} \times \dfrac{{5!}}{{3!}}\]
Now we will use the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,
\[ \Rightarrow \dfrac{{3!}}{{1!(3 - 1)!}} \times \dfrac{{2!}}{{2!(2 - 2)!}} \times \dfrac{{5!}}{{3!}}\]
Now we will simplify, then we will get,
\[ \Rightarrow \dfrac{{3!}}{{1!(2)!}} \times \dfrac{{2!}}{{2!(0)!}} \times \dfrac{{5!}}{{3!}}\]
Now we will further simplify by using factorial formula \[n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times .........1\], we get,
\[ \Rightarrow \dfrac{{3 \times 2 \times 1}}{{1 \times 2 \times 1}} \times \dfrac{{2 \times 1}}{{2 \times 1 \times 1}} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}}\]
Now we will further simplify,
\[ \Rightarrow 3 \times 1 \times 20\],
Now we will multiply, then we will get,
\[ \Rightarrow 60\],
Case II: There will be 2 pairs of 1 kind of perfume and 1 different kinds of perfume,
\[\therefore \] Number of arrangements of 2 pairs of 1 kind of perfume and 1 different kind of perfume will be:
\[ \Rightarrow {}^3{C_2} \times {}^1{C_1} \times \dfrac{{5!}}{{2! \times 3!}}\]
Now we will use the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,
\[ \Rightarrow \dfrac{{3!}}{{2!(3 - 2)!}} \times \dfrac{{1!}}{{1!(1 - 1)!}} \times \dfrac{{5!}}{{2! \times 2!}}\]
Now we will simplify, then we will get,
\[ \Rightarrow \dfrac{{3!}}{{2!(1)!}} \times \dfrac{{1!}}{{1!(0)!}} \times \dfrac{{5!}}{{2! \times 2!}}\]
Now we will further simplify by using factorial formula\[n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times .........1\], we get,
\[ \Rightarrow \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} \times \dfrac{{1 \times 1}}{{1 \times 1}} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}\]
Now we will again further simplify, then we will get,
\[ \Rightarrow 3 \times 1 \times 30\]
Now we will multiply the numbers, we will get,
\[ \Rightarrow 90\]
\[\therefore \] The total number of arrangements of perfume bottles in his showcase \[ = 60 + 90 = 150\]

The correct option is c

Note: The combination is a way of selecting elements from a set so that the order of selection doesn’t matter. With the combination, only choosing elements matters. It means the order in which elements are chosen is not essential.
A permutation is an arrangement in a definite order of several objects taken, some or all at a time, with permutations, every tiny detail matters. It means the order in which elements are arranged is significant.