Question

There are 5 green balls of different shades and 4 red balls of identical shades. No. of ways of arranging them in a row so that ·no two red balls are together is

Answer 1

1800

Answer 2

Let the number of green balls be $n_G = 5$ and the number of red balls be $n_R = 4$.
The green balls are of different shades, so they are distinguishable.
The red balls are of identical shades, so they are indistinguishable.
We want to arrange these balls in a row such that no two red balls are together.

The strategy is to first arrange the objects that are allowed to be together (the green balls) and then place the objects that cannot be together (the red balls) in the spaces created by the first set of objects.

Step 1: Arrange the 5 distinct green balls.

The number of ways to arrange 5 distinct green balls in a row is the number of permutations of 5 distinct objects, which is $5!$.
Number of ways to arrange green balls = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Step 2: Identify the spaces where the red balls can be placed.

When the 5 green balls are arranged in a row, they create spaces where the red balls can be placed so that no two red balls are adjacent. Let the positions of the green balls be represented by $G$. The spaces are before the first green ball, between any two green balls, and after the last green ball.
$\_ G \_ G \_ G \_ G \_ G \_$
There are $5+1 = 6$ such spaces.

Step 3: Place the 4 identical red balls in these spaces such that no two red balls are in the same space.

To ensure that no two red balls are together, we must place at most one red ball in each of the 6 available spaces. We have 4 identical red balls to place.
Since the red balls are identical, the number of ways to place them is determined by the number of ways to choose the 4 spaces out of the 6 available spaces.
The number of ways to choose 4 spaces out of 6 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n=6$ (number of spaces) and $k=4$ (number of red balls).
Number of ways to choose spaces for red balls = $\binom{6}{4}$.
$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{6 \times 5}{2} = \frac{30}{2} = 15$.
Once the 4 spaces are chosen, the 4 identical red balls are placed in these chosen spaces in only 1 way.

Step 4: Calculate the total number of arrangements.

The total number of ways to arrange the balls such that no two red balls are together is the product of the number of ways to arrange the green balls and the number of ways to place the red balls in the available spaces.
Total number of arrangements = (Number of ways to arrange green balls) $\times$ (Number of ways to place red balls)
Total number of arrangements = $5! \times \binom{6}{4} = 120 \times 15 = 1800$.