Question

A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair is
A. 1920
B. 200
C. 110
D. 80

Answer 1

Hint: According to the question we have to choose 4 shoes out of 5 different pairs of shoes so that there will be no complete pair. So, we use the theory of combination to solve this question.
The formula we use for combination is \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, $n=$ number of items/objects
And $r=$ number of items/objects being chosen at a time

Complete step by step answer:
We have given that a rack has 5 different pairs of shoes.
We have to find the number of ways in which 4 shoes can be chosen from it so that there will be no complete pair.
First we calculate the number of ways to choose 4 pairs out of 5 pairs, we get
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[{}^{5}{{C}_{4}}=\dfrac{5!}{4!\left( 5-4 \right)!}\]
When we solve the above equation, we get
\[\begin{align}
  & {}^{5}{{C}_{4}}=\dfrac{5\times 4!}{4!\left( 1 \right)!} \\
 & {}^{5}{{C}_{4}}=5 \\
\end{align}\]
Now, if we choose one shoe from each 4 pairs, there is no complete pair remaining.
So, the number of ways to choose a shoe from each pair will be
$\begin{align}
  & ={}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}} \\
 & =2\times 2\times 2\times 2 \\
 & =16 \\
\end{align}$
So, the total number of ways will be
$\begin{align}
  & =16\times 5 \\
 & =80 \\
\end{align}$
So, the number of ways in which 4 shoes can be chosen from it so that there will be no complete pair is $80$.
Option D is the correct answer.


Note: To solve this type of questions we must remember the formula. Also, students must know how to solve the factorial. If students apply the wrong formula, they get incorrect answers. Also, it is necessary to multiply the number of ways to choose 4 pairs out of 5 pairs with the number of ways to choose a shoe from each pair to get the required answer.