Question

In how many ways can 5 prizes be distributed to 3 students, if each student is eligible for any number of prizes? (1) \(3^{5}\) (2) \(5^{3}\) (3) \({ }^{5} \mathrm{P}_{3}\) (4) \({ }^{5} \mathrm{C}_{3}\)

Short answer

Answer: \(3^{5}\)

Step-by-step solution

Identify the type of problem

This problem involves distributing objects (prizes) into distinct boxes (students) without any restrictions. This type of problem can be solved using the formula: Number of ways = \({n^{r}}\), where \(n\) is the number of boxes (students) and \(r\) is the number of objects (prizes).

Apply the formula to the given problem

Now we need to apply this formula to our problem, where we have 5 prizes (\(r = 5\)) and 3 students (\(n = 3\)): Number of ways = \({3^{5}}\)

Identify the correct answer

From the given options: (1) \(3^{5}\) (2) \(5^{3}\) (3) \({ }^{5} \mathrm{P}_{3}\) (4) \({ }^{5} \mathrm{C}_{3}\) We can see that the correct answer is option (1), which is \(3^{5}\).

Key concepts

Distribution of Objects

In combinatorics, the distribution of objects involves allocating a certain number of items, such as prizes, into distinct containers, which could be students, teams, or any other group. In this particular exercise, we have to distribute 5 prizes among 3 students.

When dealing with the distribution of objects, it's important to consider whether each object can go into any container or if there are restrictions. Here, any student could receive any number of prizes, meaning each prize is independent of the others in terms of distribution.

The formula used when there's no restriction on how objects are distributed is \(n^r\), where \(n\) is the number of recipients (i.e., the containers) and \(r\) is the number of objects. This simple multiplication arises because each prize can be given to any of the students independently.

A real-world analogy could be imagining that each prize is a ball and each student is a basket. You can place the balls in any of the baskets freely, and you want to find out in how many different overall combinations the balls could be distributed.

Permutations

Permutations are a key concept in combinatorics. They refer to the different possible arrangements of a set of items where the order matters. Not to be confused with combinations, permutations can drastically change the outcome just by changing the sequence of items.

For example, if you were to arrange letters, "ABC" is a different permutation than "CBA". This concept is crucial when each position or recipient of an item must be unique and specific. The formula for permutations when choosing \(k\) objects from a set of \(n\) objects is given by:

• \( ^{n}P_{k} = \frac{n!}{(n-k)!} \)

It's important to note that in the context of the original exercise, permutations are not the correct approach because each prize does not need to be assigned in a unique order or to a unique student.

Combinations

Combinations are another essential component of combinatorics, focusing on the selection of objects without regard to the order. This means that, unlike permutations, "ABC" is considered the same combination as "CBA".

Combinations are used when the grouping itself matters, but the order within each group does not. The formula for combinations when choosing \(k\) objects from \(n\) possibilities is:

• \( ^{n}C_{k} = \frac{n!}{k!(n-k)!} \)

In the context of the original exercise, combinations are also not the correct solution because the problem does not ask for a unique selection of prizes per student without repetition. The correct concept here involves repetition of selections, which aligns with the formula \(n^r\) used in distributions.