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The total number of ways of selecting 10 balls out of an unlimited number of identical white, red and blue balls is equal to (1) \({ }^{12} \mathrm{C}_{2}\) (2) \({ }^{13} \mathrm{C}_{3}\) (3) \({ }^{10} \mathrm{C}_{2}\) (4) \({ }^{10} \mathrm{C}_{3}\) (5) \({ }^{12} \mathrm{C}_{3}\)

Short Answer

Expert verified
The number of ways is \( \binom{12}{2} \), so option (1) is correct.

Step by step solution

01

Understanding the problem

The problem is about finding the number of ways to select 10 balls from an unlimited supply of identical white, red, and blue balls. This is a combinatorial problem where we need to find the number of solutions to the equation.
02

Identify the equation to solve

Let the number of white balls be represented by variable w, red balls by r, and blue balls by b. We need the number of non-negative integer solutions to the equation: \[ w + r + b = 10 \]
03

Apply the stars and bars method

The stars and bars method gives the number of ways to distribute n indistinguishable objects into k distinguishable bins. In this case, it translates to finding the number of ways to partition 10 balls (stars) into 3 groups (bins) separated by 2 bars.
04

Calculate the combination

According to the stars and bars formula, the number of ways to solve the equation \( w + r + b = 10 \) is given by: \[ \binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} \]
05

Identify the correct option

Compare the result \( \binom{12}{2} \) with the given options. The correct term is (1) \( \binom{12}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Combinations
In combinatorics, combinations are a way to select items from a collection, such that the order of selection does not matter. For example, if you have a set of 3 balls, red, blue, and white, and you want to select 2 out of them, the possible combinations are: {red, blue}, {red, white}, and {blue, white}. Notice that {red, blue} is the same as {blue, red}.

This concept is crucial when dealing with problems where you need to pick a certain number of objects out of a larger set without considering the order. The general formula for combinations is given by: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n \) is the total number of items, \( r \) is the number of items to choose, and \( ! \) denotes the factorial function.

In the given exercise, we see combinations in action when selecting 10 balls from an unlimited supply of red, blue, and white balls.
Stars and Bars Method
The stars and bars method is a clever way to solve problems involving the distribution of indistinguishable objects into distinguishable bins. This method helps us find the number of ways to put \( n \) indistinguishable items into \( k \) distinguishable bins.

Imagine you have 10 stars and you need to divide them into 3 groups. You can use 2 bars to create 3 sections (one between each pair of bins). Each different arrangement of the stars and bars represents a unique distribution.

For instance, the sequence \[ **|****|**** \] means 2 items in the first bin, 4 in the second, and 4 in the third.

The total number of ways to arrange \( n \) stars and \( k - 1 \) bars is given by the combinatorial expression: \[ \binom{n + k - 1}{k - 1} \]
In the given exercise, the problem translates to finding the number of ways to place 10 stars (balls) into 3 bins (each bin representing a color), using 2 bars to separate them.
Introduction to Binomial Coefficients
Binomial coefficients are a fundamental concept in combinatorics, used to describe the coefficients in the expansion of a binomial raised to a power. They are represented by \( \binom{n}{r} \), which is read as 'n choose r'. These coefficients are key in calculating combinations and are used extensively in the stars and bars method.

The binomial coefficient \( \binom{n}{r} \) can be calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n \) represents the total number of items, and \( r \) represents the number of items being chosen. Factorials are used in this calculation, where \( n! \) (n factorial) is the product of all positive integers less than or equal to \( n \).

In the given exercise, the solution involves using the binomial coefficient to calculate \( \binom{12}{2} \), which simplifies the process of finding the number of combinations.

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