Concept not clear: Permutations and Combinations

Question

Q: There are $5$ shirts all of different colors, $4$ pairs of pants all of different colors, and $2$ pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each ? ( Question from Brilliant App ).

Solution: $(5C2 \times 2!) \times (4C2 \times 2!) \times (2C2 \times 2!) = 480$

My approach: No of permutations $= 5 \times 4 \times 2 = 40$ ways of ordering

Choosing 2 from these $40$ orderings $= 40C2 \times 2! = 1560$

Please let me know what is wrong with my approach.

Answer 1

I'd suggest to consider 2 approaches. The first one will help you to understand why your approach is not correct. Then second will help to understand how is it related to combinations and permutations.

1. The numbers are small. That's why first you can apply a straight forward approach, without combinations and permutations. How may choices of short do you have for Amy? It is 5. If you have chosen one color for Amy, Bunny can choose one of remaining colors. How many remain? 4. Thus, you have 5x4 pairs of shirt colors. Same about pants: Amy has 4 choices, for Bunny remain 3 choices. Thus there are 4x3 pants choices. For shoes Amy has 2 choices, but Bunny has no choice (i.e. only 1 choice) after Amy has chosen. That's why we have 2x1 choices of shoes. Thus the total number of ways is (5x4) x (4x3) x (2x1) = 480.

2. Now we can consider it from the point of view of combinations and permutations. We have 2 people. From 5 shirt colors we have to chose 2. How many ways are there to chose 2 out of 5? There are 5C2 ways. When we have chosen 2 shirt colors, in how many ways we can assign these colors to 2 persons? In 2! ways. If we had more colors and more persons, it would be easier to see, why we are talking about permutations here. E.g. if we had chosen 7 colors out of 20, the number of way to assign 7 colors to 7 persons would be the number of permutations, i.e. 7x6x5x4x3x2x1 = 7!. For 2 persons we have 2!. So, for shirts we have 5C2x2! For pants we have 4C2 ways to chose 2 colors out of 4. And there are 2! ways to assign them to Amy and Bunny. Thus there are 4C2x2! ways for pants. The same for shoes: 2C2x2!. The total number of ways is (5C2x2!) x (4C2x2!) x (2C2x2!) = 480.

Answer 2

Others have specifically addressed where you went wrong.

The standard way of attacking the problem is as follows:

Let

$$\binom{n}{k} = \frac{n!}{[k!][(n-k)!]} ~\text{where}~ n \in \mathbb{Z^+}, ~k \in \{0,1,2,\cdots, n\} ~\text{and}~ 0! = 1.$$

Here, $\binom{n}{k}$ represents the number of different ways of selecting $k$ items from a group of $n$ items.

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Let $A, B, C$ denote the # of different ways that Amy and Bunny can select shirts, pants, and shoes, respectively.

Then the final answer will be

$$A \times B \times C.$$

If Amy and Bunny are selecting $2$ items, from a group of $n$ items, superficially, you would suppose that the number of ways of doing this is $\binom{n}{2}$. However, this overlooks that once the two items (say one green, one red) are selected, then you have to distinguish between [Amy selecting the green item, Bunny selecting the red item] and vice-versa.

Therefore, when computing $A,B,C$, the general formula for each variable will be

$$\binom{n}{2} \times 2.$$

Thus, since the number of shirts, pants and shoes are $5,4,2$ respectively, you have that

$$A = \binom{5}{2} \times 2 = 20,$$

$$B = \binom{4}{2} \times 2 = 12,$$

$$C = \binom{2}{2} \times 2 = 2.$$

Answer 3

Your answer is completely wrong , because when you multiply $5,4,3$ , you find the possible choice for only one individual.The rest is also confusing , as far as i could understand , you try to have random $2$ combinations from among these $40$ different combination.The problem is that you try to pick up $2$ things from tne number of permutation.

Be careful!Permutation does not give you subsets, it only gives the number of ordering,so you cannot choose somethings from permutation you can make choosing only from sets.This was the key sentence for you.