Question

In how many ways can 17 billiard balls be arranged if 7 of them are black, 6 red and 4 white?

Answer 1

Hint: To solve this question, we will assume that all the balls of the same color are identical. Now, to arrange all these balls, we will apply the following formula:
$\text{Total Arrangements }=\dfrac{X!}{\left( {{x}_{1}}! \right)\left( {{x}_{2}}! \right)\left( {{x}_{3}}! \right)......\left( {{x}_{n}}! \right)}$

Complete step-by-step answer:
In the above formula, X = total number of entities, ${{x}_{1}}$ are the same kind of entities, that is they are identical, ${{x}_{2}}$ are also the same kind of entities but different from ${{x}_{1}}$. Similarly, ${{x}_{3}},{{x}_{4}},{{x}_{5}}.....{{x}_{n}}$ are the number of identical entities but different from each other.

Before solving the question, we are going to assume that all the balls of the same color are identical. This means that one red ball is identical to another red ball. Similarly, one black ball is identical to all the other black balls and one white ball will be similar to other white balls. Now, we are given that out of 17 balls, 7 of them are black, 6 are red and 4 are white. Now, we will arrange these balls in a row. The formula by which we can arrange the total number of entities which contain similar entities is given as:
$\text{Total Arrangements }=\dfrac{X!}{\left( {{x}_{1}}! \right)\left( {{x}_{2}}! \right)\left( {{x}_{3}}! \right)......\left( {{x}_{n}}! \right)}$
In this formula, X is the total number of entities, ${{x}_{1}}$ is the number of identical entities of the first kind, ${{x}_{2}}$ is the number of identical entities of the second kind, and so on. In our case, $X=17,{{x}_{1}}=7,{{x}_{2}}=6\text{ and }{{x}_{3}}=4$. Thus, we get,
$\text{Total Arrangements }=\dfrac{17!}{\left( 7! \right)\left( 6! \right)\left( 4! \right)}$
$\text{Total Arrangements }=\dfrac{17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7!}{\left( 7! \right)\left( 6! \right)\left( 4! \right)}$
$\text{Total Arrangements }=\dfrac{17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8}{\left( 6\times 5\times 4\times 3\times 2\times 1 \right)\left( 4\times 3\times 2\times 1 \right)}$
$\text{Total Arrangements }=\dfrac{17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8}{\left[ \left( 6\times 2 \right)\times \left( 5\times 3 \right)\times \left( 4\times 1 \right) \right]\times \left[ \left( 4\times 2 \right)\times \left( 3\times 1 \right) \right]}$
$\text{Total Arrangements }=\dfrac{17\times 16\times 15\times 14\times 13\times 12\times 11\times 10\times 9\times 8}{12\times 15\times 4\times 8\times 3}$
$\text{Total Arrangements }=\dfrac{17\times 16\times 14\times 13\times 11\times 10\times 9}{4\times 3}$
$\text{Total Arrangements }=\dfrac{17\times 14\times 13\times 11\times 10\times 3\times 3\times 4\times 4}{4\times 3}$
$\text{Total Arrangements }=17\times 14\times 13\times 11\times 10\times 3\times 4$
$\text{Total Arrangements }=4084080$
Thus, there are 4084080 ways in which we can arrange these billiard balls.

Note: We cannot arrange the billiards balls as follows: There are 17 balls, so the total number of arrangements = 17! This is incorrect because this method is applicable only when the balls are distinct, not identical. But, in our case, all the balls of the same color are identical.