Question

A closet has \[5\] pairs of different types of shoes. How many ways $ 4 $ shoes can be drawn from the closet so that there will be no complete pair-?
A) $ 200 $
B) $ 160 $
C) $ 40 $
D) $ 80 $

Answer 1

Hint: The above given problem is from the topic permutations and combinations. Permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen. It is an ordering of objects in a definite order. Factorial: Factorial of a positive integer $ n $ , denoted by $ n! $ , is the outcome of all positive integers \[\;\; \leqslant n\]:
 $ n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \left( {n - 3} \right) \times ........ \times 3 \times 2 \times 1 $
For example:
 $ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $
Given,
So, firstly we know the total no. of pairs given are 5.
From which we have to choose 4 shoes so that there will be no complete pair.
So, the total number of the shows $ = 10 $
We have to choose the 4 shoes but they should not be of the same pair.

 Complete step by step answer:
So, let us assume that there are four places where we have to put the shoes.
For getting the 1st shoe we can draw the shoe from 10 shoes. So, for the 1st shoe we have 10 choices, after that we are left with 9 shoes but we cannot choose the shoe of the pair from which the first shoe is drawn, so finally we are having 8 choices for the 2nd shoe which is to be drawn.
Likewise, for the 3rd show there will be 6 choices and for the 4th shoe there will be 4 choices.

If we take it in an orderly form, the possibility may be,
10,8,6,4.
So, the possibilities to draw 4 shoes of pair is $ = 10 \times 8 \times 6 \times 4 = 1940 $
But it is not mentioned that we have to arrange it in an orderly form. We have to choose 4 shoes which means we have 4 slots to fill by selecting shoes from 5 pair of shoes ,we can put the shoe selected in any slot out of 4 be it the first , second, third or fourth .So, we have 4 ways to fill at first then 3 ways and so on which makes the possibility of arrangement as follows:

We need to reduce the possibilities of the arrangement of 4 slots from the possibility of selection of 4 shoes out of 10:
= $ \dfrac{{1940}}{{4!}} $
= $ \dfrac{{10 \times 8 \times 6 \times 4}}{{4 \times 3 \times 2 \times 1}} $
= $ 80 $
We have one more method to solve this question which is performed below -
Total shoes $ \Rightarrow 5 $ pair of shoes
=Total ways of selecting $ 4 $ pair out of $ 5 $
 $ \Rightarrow {}^5{C_4} $
Now for every pair, on selecting $ 1 $ shoe,
This can be done in $ {\left( {{}^2{C_1}} \right)^4} $ ways
Since, Total no. of ways $ = \left( {{}^5{C_4}} \right){\left( {{}^2{C_1}} \right)^4} $
 $ = \left( {\dfrac{{5 \times 4!}}{{4!}}} \right) \times {\left( {\dfrac{{2 \times 1}}{{1!}}} \right)^4} $
 $ = 5 \times {2^4} $
 $ = 5 \times 16 $
 $ = 80 $
Hence, the correct answer is option D.

Note:
The calculation of permutation and combinations should not be confused because of factorial. Remember, $ 0! = 1 $ . For any number n, $ n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \left( {n - 3} \right) \times ........ \times 3 \times 2 \times 1 $ For example: $ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $ . So, in a closet has $ 5 $ pairs of different types of shoes. We have to choose $ 4 $ shoes from the
closet so that there will be no complete pair is-
10,8,6,4
 $ \Rightarrow 10 \times 8 \times 6 \times 4 = 1940 $
If the possibility is not in the orderly form, (knowledge required- $ \left( {4! = 4 \times 3 \times 2 \times 1} \right) $ )
 $ = \dfrac{{1940}}{{4!}} $
 $ = \dfrac{{10 \times 8 \times 6 \times 4}}{{4 \times 3 \times 2 \times 1}} $
 $ = 80 $ .
So, this is another method to get the solution.