The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remain

Q: Wait, why is the number of ways to choose 'r' identical objects always 1?

Think about it this way: if you have 10 identical blue marbles and I ask you to pick 3, it doesn't matter which 3 you pick because they are indistinguishable. Any selection of 3 marbles looks exactly the same as any other selection of 3 marbles. Therefore, there is only 1 unique way to choose 'r' identical objects.

Q: Why does the sum go from r=0 to 10?

We need to select a total of 10 objects. We can choose anywhere from 0 to 10 identical objects. If we choose 0 identical objects, we must choose 10 distinct ones. If we choose 10 identical objects, we must choose 0 distinct ones. Thus, 'r' can range from 0 to 10, covering all possible combinations.

Q: Why is the sum of $\binom{21}{0} + \dots + \binom{21}{10}$ equal to half of $2^{21}$?

This is a beautiful application of the symmetry property $\binom{n}{k} = \binom{n}{n-k}$. The total sum of all binomial coefficients $\sum_{k=0}^{21} \binom{21}{k}$ is $2^{21}$. Because $\binom{21}{0} = \binom{21}{21}$, $\binom{21}{1} = \binom{21}{20}$, and so on, the first half of the sum (from $k=0$ to $10$) is exactly equal to the second half (from $k=11$ to $21$). Since the two halves are equal and their total is $2^{21}$, each half must be $2^{21} / 2 = 2^{20}$.

Question

The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :

Options:

(A) $2^{20}$

(B) $2^{20} - 1$

(C) $2^{20} + 1$

(D) $2^{21}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with selecting exactly $10$ objects from a collection of $31$ items. This collection consists of $10$ identical objects and $21$ distinct objects.

To solve this, we define $r$ as the number of identical objects chosen. Since there are $10$ identical objects available, $r$ can range from $0$ to $10$.

### The Case-by-Case Strategy

For any chosen value of $r$, we must select the remaining $10 - r$ objects from the $21$ distinct objects.

Because the $10$ identical objects are indistinguishable, there is only $1$ way to choose $r$ of them. The number of ways to choose the distinct objects is given by the combination formula $\binom{21}{10-r}$.

Thus, for a specific $r$, the number of ways to form the selection is:
$$1 	imes \binom{21}{10-r} = \binom{21}{10-r}$$

### The Summation Challenge

To find the total number of ways, $S$, we sum these possibilities across all valid values of $r$:
$$S = \sum_{r=0}^{10} \binom{21}{10-r}$$

Expanding this sum, we obtain:
$$S = \binom{21}{10} + \binom{21}{9} + \dots + \binom{21}{0}$$

We know that the sum of all binomial coefficients for a given $n$ is $2^n$. Specifically, for $n=21$:
$$\sum_{k=0}^{21} \binom{21}{k} = 2^{21}$$

### The Symmetry Breakthrough

We utilize the symmetry property of binomial coefficients, which states that $\binom{n}{k} = \binom{n}{n-k}$. This implies that $\binom{21}{0} = \binom{21}{21}$, $\binom{21}{1} = \binom{21}{20}$, and generally $\binom{21}{k} = \binom{21}{21-k}$.

Because $21$ is odd, the total sum $\sum_{k=0}^{21} \binom{21}{k}$ consists of $22$ terms, which can be split into two equal halves of $11$ terms each. The first half is $\sum_{k=0}^{10} \binom{21}{k}$, which is exactly our sum $S$.

The second half is $\sum_{k=11}^{21} \binom{21}{k}$, which is also equal to $S$ due to the symmetry property. Therefore, we can write:
$$S + S = 2^{21}$$

### Final Calculation

Simplifying the equation above, we get:
$$2S = 2^{21}$$

Dividing both sides by $2$, we arrive at the final result:
$$S = 2^{20}$$

The total number of ways to select the objects is **$2^{20}$**.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :

Select Answer:

Visualized Solution

Analyze the Object Composition

Selection Strategy

Ways to Choose Identical Objects

Ways to Choose Distinct Objects

Total Ways for a Specific $r$

Summing Over All Possible $r$

Binomial Theorem Identity

Symmetry of Binomial Coefficients

Splitting the Total Sum

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Case-by-Case Strategy

The Summation Challenge

The Symmetry Breakthrough

Final Calculation

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Animated Solution for Mathematics - Permutations and Combinations: The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is :

Select Answer:

Visualized Solution

Analyze the Object Composition

Selection Strategy

Ways to Choose Identical Objects

Ways to Choose Distinct Objects

Total Ways for a Specific .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Summing Over All Possible .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Binomial Theorem Identity

Symmetry of Binomial Coefficients

Splitting the Total Sum

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Case-by-Case Strategy

The Summation Challenge

The Symmetry Breakthrough

Final Calculation

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