Question

There are four different white balls and four different black balls. The number of ways that balls can be arranged in a row so that white and black balls are placed alternately is (1) \((4 !)^{2}\) (2) \(2(4 !)^{2}\) (3) \(4 !\) \(\begin{array}{ll}(4) & (4 !)^{3}\end{array}\)

Short answer

Answer: \(2(4!)^2\) ways.

Step-by-step solution

Identify the possible starting points

There are two possibilities for starting the row of alternating balls: either starting with a white ball or starting with a black ball. In either case, we'll need to arrange the balls in the row, keeping them in an alternating pattern. Notice that the total number of balls is the same in either case (8 total balls).

Count the possible arrangements starting with a white ball

If the row starts with a white ball, we can imagine placing 4 white balls in the row first, followed by filling the gaps with the 4 black balls. To place the white balls, consider that there are initially 4 empty spots. We have 4 choices for the first white ball, 3 choices for the second white ball, 2 choices for the third white ball, and 1 choice for the last white ball. Thus, there are 4! ways to place the white balls. Next, we place the black balls in the gaps between the white balls. There are 4 gaps (one in between each pair of neighboring white balls, plus one gap after the last white ball). So we have 4 choices for the first black ball, then 3 choices, 2 choices, and finally 1 choice. There are 4! ways to place the black balls. Since we need to place both types of balls, multiply their combinations: (4!) * (4!)

Count the possible arrangements starting with a black ball

Similarly to Step 2, if the row starts with a black ball, we can imagine placing the 4 black balls in the row first, followed by filling the gaps with the 4 white balls. The logic is the same as in Step 2, so there are (4!) * (4!) ways to place the balls when starting with a black ball.

Combine both possibilities

Now, we simply need to account for both starting points (starting with a white ball or a black ball). To do this, we can add the number of possible arrangements from Steps 2 and 3. So, the total number of ways to arrange the balls is: \(2(4!)^2\) Therefore, the correct answer is (2).

Key concepts

Permutations

Permutations are all about arranging items in an ordered sequence. Imagine you have a set of items, like balls of different colors, and you want to know how many different ways you can arrange them in a line. This is where permutations come into play.

The concept of permutations is fundamental in combinatorics. For example, if you have four distinct white balls, the number of ways to arrange them is given by the permutation formula, which is the factorial of the number of items, written as \(4!\).

When discussing permutations, remember:

• Order matters: changing the order creates a different sequence.
• Each unique arrangement is a permutation.

In our example, if you need to arrange both black and white balls, each group of balls can be arranged independently through permutations. Thus, for four black balls, it's also \(4!\) arrangements.

Alternating Sequences

Alternating sequences refer to a specific order where items alternate between two types or categories. In the context of our exercise, we're arranging white and black balls such that they always alternate.

There are two possible ways to start an alternating sequence: either with a white ball or a black ball first. If we start with a white ball, the sequence alternates like this: White, Black, White, Black, and so on. Conversely, if we start with a black ball, we alternate starting with Black, White, Black, White, etc.

Key points about alternating sequences:

• The sequence maintains a specific order by alternating between two distinct types.
• In combinatorics problems, determining the number of ways to create alternating sequences involves calculating permutations for each group and considering possible starting choices.

In the exercise, we observed that the alternating sequence could start with either type of ball, doubling the number of possible arrangements.

Factorials

Factorials are a way to describe a product of all positive integers up to a certain number. Factorial notation is represented with an exclamation mark, such as \(n!\).

Here's what you need to know about factorials:

• They represent permutations: the number of ways to arrange \(n\) distinct items.
• The formula is: \(n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1\).

In our exercise, factorials are crucial to calculate how many ways 4 white or 4 black balls can be arranged, which is \(4!\). This results in 24 different arrangements.

By understanding factorials, you can solve many combinatorial problems. The arrangement of elements, like our colored balls, often leans heavily on factorial calculations.