In how many ways can 21 identical white balls and 19 identical black balls be arranged in a row so that no 2 black balls are together?

Options

A. 1540
B. 1640
C. 1240
D. 1440

Answer

To solve the problem of arranging 21 identical white balls and 19 identical black balls in a row such that no two black balls are together, we can follow these steps:

### Step 1: Understand the arrangement
We have 21 identical white balls and 19 identical black balls. The condition that no two black balls can be together means that we need to place the black balls in such a way that they are separated by at least one white ball.

### Step 2: Create gaps for black balls
When we arrange the 21 white balls in a row, they create gaps where the black balls can be placed. Specifically, if we have 21 white balls, we can visualize them as follows:

W W W W W W W W W W W W W W W W W W W W W

This arrangement creates gaps for the black balls:
- One gap before the first white ball
- One gap between each pair of white balls
- One gap after the last white ball

Thus, the total number of gaps created is:
- Number of gaps = Number of white balls + 1 = 21 + 1 = 22 gaps

### Step 3: Place the black balls in the gaps
Now, we need to place the 19 identical black balls into these 22 gaps. Since no two black balls can be together, we can choose any 19 of these 22 gaps to place a black ball.

### Step 4: Calculate the combinations
The number of ways to choose 19 gaps from 22 is given by the combination formula:

\[ \text{Number of ways} = \binom{22}{19} \]

Using the combination formula, we can rewrite this as:

\[ \binom{22}{19} = \binom{22}{3} \]

### Step 5: Calculate \(\binom{22}{3}\)
Now we can calculate \(\binom{22}{3}\):

\[ \binom{22}{3} = \frac{22!}{3!(22-3)!} = \frac{22!}{3! \cdot 19!} \]

Calculating this step-by-step:

1. Calculate \(22!\) as \(22 \times 21 \times 20 \times 19!\)
2. Substitute into the formula:

\[ \binom{22}{3} = \frac{22 \times 21 \times 20 \times 19!}{3! \times 19!} \]

3. Cancel \(19!\):

\[ = \frac{22 \times 21 \times 20}{3!} \]

4. Calculate \(3! = 6\):

\[ = \frac{22 \times 21 \times 20}{6} \]

5. Now perform the multiplication:

\[ = \frac{9240}{6} = 1540 \]

### Final Answer
Thus, the total number of ways to arrange 21 identical white balls and 19 identical black balls in a row such that no two black balls are together is **1540**.

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