Question

Three identical red balls, three identical green balls and three identical blue balls are placed in a row in a random order. Let \(A\) denotes the event that no two blue balls are adjacent. The odds against ' \(A\) ' is (1) \(7: 4\) (2) \(5: 3\) (3) \(7: 5\) (4) \(6: 5\) (5) none of these

Short answer

(3) \(7:5\)

Step-by-step solution

- Determine the total number of arrangements

To find the total number of ways to arrange 9 balls (3 red, 3 green, 3 blue), we use the multinomial coefficient formula: \[ \binom{9}{3, 3, 3} = \frac{9!}{3! \times 3! \times 3!} = 1680 \]

- Understanding the restriction

To count the arrangements where no two blue balls are adjacent, consider that no two blue balls can be next to each other.

- Arrangement of red and green balls

First, arrange the 6 non-blue balls (3 red and 3 green). There are \[ \binom{6}{3, 3} = \frac{6!}{3! \times 3!} = 20 \] ways to arrange these balls.

- Insert blue balls

To ensure no two blue balls are adjacent, think of placing the blue balls into the gaps created by the 6 non-blue balls (including the ends). There are 7 such gaps: \[ - R - R - R - G - G - G - \]

- Choose gaps for blue balls

Select 3 out of these 7 gaps to place the blue balls, ensuring they are not adjacent: \[ \binom{7}{3} = 35 \]

- Total valid arrangements

Multiply the number of ways to arrange the red and green balls by the number of ways to choose the gaps for blue balls: \[ 20 \times 35 = 700 \]

- Calculate the odds

Odds against A is the ratio of unsuccessful outcomes to successful ones. There are 1680 total arrangements and 700 successful ones, so: \[ \text{Odds against A} = \frac{1680 - 700}{700} = \frac{980}{700} = \frac{7}{5} \]

Key concepts

multinomial coefficient

The multinomial coefficient is a generalization of the binomial coefficient. It is used to find the number of ways to distribute a set of items into different groups. In our problem, we need to arrange 9 balls with 3 different colors: red, green, and blue. The formula to calculate the multinomial coefficient is \(\binom{n}{k_1, k_2, \textellipsis, k_m} = \frac{n!}{k_1! k_2! \textellipsis k_m!}\) where \(!\) stands for factorial. For this exercise, we calculate \(\binom{9}{3, 3, 3} = \frac{9!}{3! \times 3! \times 3!} = 1680\). This tells us there are 1680 possible ways to arrange the balls without considering any restrictions.

combinatorial restrictions

Combinatorial restrictions limit the arrangements of items based on specific rules or conditions. In our exercise, the restriction is that no two blue balls are adjacent. We start by arranging the red and green balls first. There are 6 non-blue balls (3 red and 3 green), which can be arranged in \(\binom{6}{3, 3} = \frac{6!}{3! \times 3!} = 20\) ways. Next, we need to place the blue balls in such a way that none are adjacent. This involves placing blue balls in the gaps created by the arrangement of red and green balls. There are 7 potential gaps to place the blue balls: before, between, and after the non-blue balls. We choose 3 out of these 7 gaps, giving us \(\binom{7}{3} = 35\). Thus, the number of valid arrangements with the given restriction is \(20 \times 35 = 700\).

odds calculation

Odds calculation helps us understand the likelihood of an event happening compared to it not happening. Here, we need to calculate the odds against the event 'A' where no two blue balls are adjacent. We already know there are 1680 total possible arrangements and 700 arrangements that satisfy our condition (event A). The number of unfavorable outcomes is \(1680 - 700 = 980\). Therefore, the odds against 'A' are given by the ratio of unsuccessful outcomes to successful ones: \( \frac{980}{700} = \frac{7}{5}\). This means there are 7 unfavorable outcomes for every 5 favorable ones, which we can express as '7 to 5' or \(7:5\).