Question

There are five different teacups, three saucers and four teaspoons in the “tea party” store. How many ways are there to buy two items with different names?

Answer 1

Hint: We will solve this question by using the combination rule. A combination is the choice of $r$ things from a set of $n$ things without replacement. The order does not matter in combination${{\Rightarrow }{}^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ , where the factorial means it is a function that multiplies a number by every number below it. For example, $5!=5\times 4\times 3\times 2\times 1$.

Complete step-by-step solution:
Here we will use combinations, according to the question we have to make two ways to buy different items.
We have given$5$ different tea cups, $3$ saucers,$4$ teaspoons.
The different ways to buy $2$ items are;
 First we will take one tea cup and one saucer then the combination becomes:
$\Rightarrow {}^{5}{{C}_{1}}\times {}^{3}{{C}_{1}}$
Then we will take one tea cup and one teaspoon then the combination becomes;
$\Rightarrow {}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}$
Similarly we will take one saucer and one teaspoon then, we get
$\Rightarrow {}^{3}C{}_{1}\times {}^{4}C{}_{1}$
Now we will calculate them and then add them, we get
$\Rightarrow {}^{5}{{C}_{1}}\times {}^{3}{{C}_{1}}+{}^{5}C{}_{1}\times {}^{4}C{}_{1}{{+}^{3}}{{C}_{1}}{{\times }^{4}}{{C}_{1}}$
Now by calculating then we get the number of different ways,
$\begin{align}
  & \Rightarrow \left( 5\times 3 \right)+\left( 5\times 4 \right)+\left( 3\times 4 \right) \\
 & \Rightarrow 15+20+12 \\
 & \Rightarrow 47 \\
\end{align}$
Hence we get $47$ different ways by which we buy $2$ different items.

Note: Most probably we made a mistake in the place of multiplication we put addition, and this is the very basic mistake. Sometimes in these types of questions we make mistakes by using permutation except combination. Always be remembering in permutation order matters but in combination order doesn't matter.