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
(E) 75

Correct Answer: D
Solution and Explanation:
Approach Solution 1:
The problem statement states that:
Given:

• A rack has 5 different pairs of shoes.

Find out:

• The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair.

As per the condition of the question, we need to choose any 4 pairs out of those 5 pairs.
Therefore, we get:
5C4 = 5
From each pair selected we have 2 ways to select 1 shoe
2*2*2*2
Hence, the total number of ways = 5 * 2 * 2 * 2 * 2 = 80
Therefore, the number of ways in which 4 shoes can be chosen from it so that there will be no complete pair = 80.

Approach Solution 2:
The problem statement informs that:
Given:

• A rack has 5 different pairs of shoes.

Find out:

• The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair.

We can use the theory of permutation and combination to solve this question.
So, before cracking this question, we need to first recall the basics of this chapter.
For instance, if we need to pick two racks out of four racks, this can be done in 4C2 = 6 ways.
The formula to be used is as follows:
nCr = \(n!\over[r!(n - r)!]\)
As per the conditions of the question, 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.
Therefore, we have to use "combinations"
The formula for combination is given as: nCr = \(n!\over[r!(n - r)!]\) where n! means the factorial of n
Firstly, we must pick 4 racks out of 5 which can be chosen by:
5C4 = \(5!\over[4!(5 - 4)!]\)
= \(5×4×3×2×1\over4×3×2×1\)
= 5
Now for every pair, on selecting 1 shoe, this can be done in \((2C1)^4\)ways.
Suppose we choose left shoes only.
There will be 5 variants to pick 4 from 5.
Now we add the chance of selecting the right shoe.
Then from each rack, we have to choose 1 shoe.
Therefore, out of 2, we have to pick one which can be done as follows
= 2C1 × 2C1 × 2C1 × 2C1
= 16
Total number of ways of choosing are = 16 × 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 = 80.

Approach Solution 3:
The problem statement implies that:
Given:

• A rack has 5 different pairs of shoes.

Find out:

• The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair.

We have to select 4 shoes out of 10 since there are 5 pairs.
Pick the first shoe. 10 different ways.
Pick the second shoe: 9 different ways, but one of them would complete a pair, so 8 distinct ways to do it.
Pick the third shoe: 8 different ways, two of them would complete a pair, so 6 different ways not completing a pair.
Pick the fourth shoe: 7 different ways, or 4 different ways without completing a pair.
Therefore, there are 10×8×6×4 = 1920 different ways, if you calculate the order in which you choose.
However, the order is not important, each set of 4 different shoes, can be brought in
4! different ways.
Therefore, the answer is \({10×8×6×4\over4×3×2×1} = {1920\over24} = 80\)
Hence, the number of ways in which 4 shoes can be chosen from it so that there will be no complete pair = 80.

“A rack has 5 different pairs of shoes. The number of ways in which 4” - is a topic of the GMAT Quantitative reasoning section of the GMAT exam. The candidates must know the basic concepts of mathematics in order to solve GMAT Problem Solving questions. The candidates can follow GMAT Quant practice papers to explore varieties of questions that will enable them to attain better scores in the exam.

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