Question

Number of ways of selecting 6 shoes out of 8 pair of shoes, having exactly two pairs is

& A.\text{ }1680 \\\ & B.\text{ }240 \\\ & C.\text{ }120 \\\ & D.\text{ }3360 \\\ \end{aligned}$$

Answer 1

In this question, we will use a combination method to get our required answer. Here, we are given a total of 8 pairs of shoes and we have to choose 6 socks, but there should be only two pairs and hence, remaining two socks should not form a pair. We will consider all the possibilities and ways and then multiply them to get our final answer. Formula of combination which we will use here is given as:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Where r items are to be selected from n number of total items.

Complete step-by-step answer:
Let us first understand the meaning of combination.
Combination is a way of selecting items from a collection such that order of selection does not matter.
Let us now analyze the question carefully.
Here, we are given 8 pairs of shoes and we have to select 6 shoes out of which there should be two pairs only.
Let us first find ways of selecting 2 pair of shoes out of 8 pair of shoes which will be given by $\Rightarrow {}^{8}{{C}_{2}}=\dfrac{8!}{2!\left( 8-2 \right)!}=\dfrac{5!}{2!6!}=\dfrac{8\times 7\times 6!}{2\times 6!}=28\text{ ways}$
Hence, there are 28 ways of selecting 2 pairs of shoes.
Since, we have selected four shoes in total out of 8 pairs. Therefore, now we have to select 2 shoes out of the remaining 6 pairs of shoes which should be from different pairs. To understand this clearly, let's name ${{A}_{1}},{{A}_{2}}$ as the first pair of shoes. ${{B}_{1}},{{B}_{2}}$ as second pair of shoes and similarly ${{C}_{1}},{{C}_{2}};{{D}_{1}},{{D}_{2}};{{E}_{1}},{{E}_{2}};{{F}_{1}},{{F}_{2}}$ as all pairs where ${{A}_{1}},{{B}_{1}},{{C}_{1}},{{D}_{1}},{{E}_{1}},{{F}_{1}}$ represent left foot and ${{A}_{2}},{{B}_{2}},{{C}_{2}},{{D}_{2}},{{E}_{2}},{{F}_{2}}$ represent right foot. Now, let us select one shoe from left foot shoes. Hence, we can select one shoe out of 6 possibilities.
Number of ways of selecting one more shoe ${}^{6}{{C}_{1}}=\dfrac{6!}{1!\left( 6-1 \right)!}=\dfrac{6!}{5!}=\dfrac{6\times 5!}{5!}=6$
Now, let us consider we have ${{B}_{1}}$ as our shoe, so we cannot take ${{B}_{2}}$ as our other shoe otherwise it will form a pair and we will have three pairs but we need two pairs only. Thus, we are left with 10 choices from remaining shoes.

${{A}_{1}}$	${{A}_{2}}$	Right
${{B}_{1}}$	${{B}_{2}}$	Wrong
${{C}_{1}}$	${{C}_{2}}$	Right
${{D}_{1}}$	${{D}_{2}}$	Right
${{E}_{1}}$	${{E}_{2}}$	Right
${{F}_{1}}$	${{F}_{2}}$	Right

(10 shoes left)
Therefore, the number of ways of selecting the last shoe from the remaining 10 shoes will be ${}^{10}{{C}_{1}}=\dfrac{10!}{1!\left( 10-1 \right)!}=\dfrac{10!}{9!}=\dfrac{10\times 9!}{9!}=10\text{ ways}$
Hence, combining all possibilities, total ways become $\Rightarrow 28\times 6\times 10=1680$

So, the correct answer is “Option A”.

Note: Students should try to draw small diagrams for better understanding. They should consider all possibilities and then multiply all the ways to get the required answer. Don't get confused with multiplication and addition. Multiplication is used for 'and' and addition is used for 'or'. Students can also remember that ${}^{n}{{C}_{1}}=n$ always and need not solve it fully every time.