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: In this question we use the theory of permutation and combination. So, before solving this question you need to first recall the basics of this chapter. For example, if we need to select two racks out of four racks. in this case, this can be done in${}^{\text{4}}{{\text{C}}_2}$ =6 ways.

Formula use- ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$

Complete step-by-step answer:
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. So, we have to use "combinations"
The formula for combination is given as:
                                                         ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n
First of all, we have to choose 4 racks out of 5 which can be chosen by:
${}^5{{\text{C}}_4}{\text{ = }}\dfrac{{5!}}{{{\text{[4}}!{\text{(5 - 4)}}!{\text{]}}}}$
= $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}}$
= $5$
Now for every pair, on selecting 1 shoe,
this can be done in ${\left( {{}^{\text{2}}{{\text{C}}_{\text{1}}}} \right)^{\text{4}}}$ways.
Suppose we pick left shoes only.
There will be 5 variants to pick 4 from 5. Now we add the possibility of picking the right shoe.
Then from each rack we have to choose 1 shoe,
So out of 2 we have to choose one which can be done as follows
= ${}^{\text{2}}{{\text{C}}_{\text{1}}}$$ \times $${}^{\text{2}}{{\text{C}}_{\text{1}}}$$ \times $${}^{\text{2}}{{\text{C}}_{\text{1}}}$$ \times $${}^{\text{2}}{{\text{C}}_{\text{1}}}$
= 16
Total number of ways of choosing are = 16 $ \times $5 = 80 ways
Therefore, the number of ways in which 4 shoes can be chosen from it. so that there will be no complete pair, is 80.
Thus, option (D) is the correct answer.

Note: We need to remember this formula for selecting r things out of n.
                                                       ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n.
for example, $3!{\text{ = 3}} \times {\text{2}} \times 1$