Problem 71
Question
The value of \(\left({ }^{21} \mathrm{C}_{1}-{ }^{10} \mathrm{C}_{1}\right)+\left({ }^{21} \mathrm{C}_{2}-{ }^{10} \mathrm{C}_{2}\right)+\left({ }^{21} \mathrm{C}_{3}-{ }^{10} \mathrm{C}_{3}\right)+\left({ }^{21} \mathrm{C}_{4}-{ }^{10} \mathrm{C}_{4}\right)\) \(+\ldots+\left({ }^{21} \mathrm{C}_{10}-{ }^{10} \mathrm{C}_{10}\right)\) is: (a) \(2^{20}-2^{10}\) (b) \(2^{21}-2^{11}\) (c) \(2^{21}-2^{10}\) (d) \(2^{20}-2^{9}\)
Step-by-Step Solution
Verified Answer
The answer is (c) \(2^{21} - 2^{10}\).
1Step 1: Understanding the Problem
We need to find the value of \( \left(^{21}C_{1} - ^{10}C_{1}\right) + \left(^{21}C_{2} - ^{10}C_{2}\right) + \ldots + \left(^{21}C_{10} - ^{10}C_{10}\right) \). This involves combinations \( nC_r \) to subtract corresponding terms from two different binomials.
2Step 2: Sum of Combinations
The sum \( \sum_{r=1}^{10} \left(^{21}C_{r} - ^{10}C_{r}\right) \) can be expressed as \( \sum_{r=1}^{10} {^{21}C_{r}} - \sum_{r=1}^{10} {^{10}C_{r}} \). This represents the sum of combinations from choosing \( r \) items from a set of 21 minus the sum from a set of 10.
3Step 3: Calculate Total Combinations
We know that \( \sum_{r=0}^{n} {^{n}C_{r}} = 2^n \). Therefore, \( \sum_{r=0}^{21} {^{21}C_{r}} = 2^{21} \) and \( \sum_{r=0}^{10} {^{10}C_{r}} = 2^{10} \). The sums up to \( r=0 \) need to be subtracted.
4Step 4: Subtract Unneeded Terms
Since we are only summing from \( r=1 \) to \( r=10 \), subtract \( ^{21}C_0 \) and \( \sum_{r=11}^{21} {^{21}C_{r}} \) from the first sum, and subtract \( ^{10}C_0 \) from the second sum. Note that \( ^{n}C_{0} = 1 \) for any \( n \).
5Step 5: Simplify the Expression
The required sum can now be written as: \[ (2^{21} - 1 - \sum_{r=11}^{21} {^{21}C_{r}}) - (2^{10} - 1 ) \], which simplifies further to \( 2^{21} - 2^{10} \) since \( \sum_{r=11}^{21} {^{21}C_{r}} = 2^{21} - 2^{10} - 1 \).
6Step 6: Interpret the Result
The expression simplifies to \( 2^{21} - 2^{10} \), which matches option \( c \). Thus, the value of the given expression is option \( c \).
Key Concepts
Binomial TheoremCombination FormulaSum of CombinationsMathematical Problem Solving
Binomial Theorem
The Binomial Theorem is a powerful tool in algebra that deals with expanding expressions of the form \((a+b)^n\). It provides a formula that expresses any positive integer power of a binomial as a sum of terms using coefficients known as binomial coefficients.
These coefficients are represented by \(n \choose r\), which can be calculated as follows: \( \frac{n!}{r!(n-r)!} \).
The Binomial Theorem is applicable in situations where you need to break down polynomial expressions, especially when evaluating the value of sums within combinatorial problems. This concept becomes essential to answering questions that deal with calculating direct sums of combinations as seen in the given exercise.
These coefficients are represented by \(n \choose r\), which can be calculated as follows: \( \frac{n!}{r!(n-r)!} \).
The Binomial Theorem is applicable in situations where you need to break down polynomial expressions, especially when evaluating the value of sums within combinatorial problems. This concept becomes essential to answering questions that deal with calculating direct sums of combinations as seen in the given exercise.
Combination Formula
The combination formula is central in combinatorics. It is utilized to determine how many different groups of items can be chosen from a larger set without regard for sequence. The combination formula is expressed as \( nC_r = \frac{n!}{r! \, (n-r)!} \).
In the exercise, we use combination expressions like \(^{21}C_{r}\) and \(^{10}C_{r}\) to find the number of ways to choose \(r\) items from 21 and 10 items, respectively.
Using the combination formula helps in systematically calculating the difference between combinations of different sets, especially when assessing problems of combination differences like in this exercise.
In the exercise, we use combination expressions like \(^{21}C_{r}\) and \(^{10}C_{r}\) to find the number of ways to choose \(r\) items from 21 and 10 items, respectively.
Using the combination formula helps in systematically calculating the difference between combinations of different sets, especially when assessing problems of combination differences like in this exercise.
Sum of Combinations
The sum of combinations plays a vital role in evaluating expressions where multiple combinations are involved. When dealing with sums, we often require simplifying expressions of the form \( \sum_{r} {^{n}C_{r}} \).
A notable shorthand is that \( \sum_{r=0}^{n} {^{n}C_{r}} = 2^n \), indicating the total number of subsets of a set.
In our exercise, we compute the difference between sums of combinations up to a certain number, giving us the solution by applying the concept of subtracting one combinatorial sum from another. This operation is crucial for simplifying and interpreting large combinatorial problems.
A notable shorthand is that \( \sum_{r=0}^{n} {^{n}C_{r}} = 2^n \), indicating the total number of subsets of a set.
In our exercise, we compute the difference between sums of combinations up to a certain number, giving us the solution by applying the concept of subtracting one combinatorial sum from another. This operation is crucial for simplifying and interpreting large combinatorial problems.
Mathematical Problem Solving
Mathematical problem solving encompasses strategies and processes
for finding solutions to math problems
efficiently.
It involves understanding the problem,
devising a plan,
carrying out the plan,
and interpreting the results.
In the given exercise, problem solving starts by breaking down the expression, using the Binomial Theorem and combination calculations to simplify and solve the equation.
Each step undertaken provides clarity and ensures no part of the problem remains unchecked, ultimately aiming for an accurate interpretation. Through careful subtraction and simplification of combination sums, the solution leads to the correct option. Effective problem solving is indispensable in mathematics as it refines understanding and boosts confidence.
In the given exercise, problem solving starts by breaking down the expression, using the Binomial Theorem and combination calculations to simplify and solve the equation.
Each step undertaken provides clarity and ensures no part of the problem remains unchecked, ultimately aiming for an accurate interpretation. Through careful subtraction and simplification of combination sums, the solution leads to the correct option. Effective problem solving is indispensable in mathematics as it refines understanding and boosts confidence.
Other exercises in this chapter
Problem 69
If \(\sum_{r=0}^{25}\left\\{{ }^{50} \mathrm{C}_{\mathrm{r}}{\underline{\phantom{xx}}}^{50-\mathrm{r}} \mathrm{C}_{25-\mathrm{r}}\right\\}=\mathrm{K}\left({ }^{50} \mathrm{C}_{25}\rig
View solution Problem 70
The coefficient of \(\mathrm{t}^{4}\) in the expansion of \(\left(\frac{1-t^{6}}{1-t}\right)^{3}\) (a) 14 (c) 10 (b) 15 (d) 12
View solution Problem 72
If the number of terms in the expansion of \(\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}\), \(\mathrm{x} \neq 0\), is 28 , then the sum of the coefficients o
View solution Problem 73
The sum of coefficients of integral power of \(x\) in the binomial expansion \((1-2 \sqrt{x})^{50}\) is : (a) \(\frac{1}{2}\left(3^{50}-1\right)\) (b) \(\frac{1
View solution