Problem 47
Question
If \(S=\sum_{r=0}^{m}{\underline{\phantom{xx}}}^{n+r} C_{k}\), then (A) \(S+{ }^{n} C_{k+1}={ }^{n+m} C_{k+1}\) (B) \(S+{ }^{n} C_{k+1}={ }^{n+m+1} C_{k+1}\) (C) \(S+{ }^{n} C_{k}={ }^{n+m} C_{k}\) (D) None of these
Step-by-Step Solution
Verified Answer
B. \(S + \binom{n}{k+1} = \binom{n+m+1}{k+1}\).
1Step 1: Recognize the Summation
The given sum is \(S = \sum_{r=0}^{m} \binom{n+r}{k}\). We need to find which equation satisfies this sum.
2Step 2: Use Identity for Summation of Combinations
A useful identity here is \(\sum_{r=0}^{m} \binom{n+r}{k} = \binom{n+m+1}{k+1} - \binom{n}{k+1}\), which comes from the hockey-stick identity for combinations.
3Step 3: Add \(\binom{n}{k+1}\) to Both Sides
To solve for \(S + \binom{n}{k+1}\), add \(\binom{n}{k+1}\) to the right-hand side of the identity: \(\sum_{r=0}^{m} \binom{n+r}{k} + \binom{n}{k+1} = \binom{n+m+1}{k+1}\).
4Step 4: Identify the Correct Option
The expression now is \(S + \binom{n}{k+1} = \binom{n+m+1}{k+1}\). This matches with option B in the provided list of options.
Key Concepts
binomial coefficientshockey-stick identitysummation of combinations
binomial coefficients
One of the cornerstones of combinatorics, binomial coefficients, are numbers that describe the number of ways to choose a subset of items from a larger set. This is typically represented as \( \binom{n}{k} \), which is read as "n choose k". This denotes the number of ways to choose \( k \) elements from a set of \( n \) elements, disregarding the order of selection.
The formula to compute a binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( ! \) denotes factorial, the product of all positive integers up to that number. Binomial coefficients are useful for solving problems related to combinations and permutations, as they provide a quick way to count combinations.
The formula to compute a binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( ! \) denotes factorial, the product of all positive integers up to that number. Binomial coefficients are useful for solving problems related to combinations and permutations, as they provide a quick way to count combinations.
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \)
- \( \binom{10}{3} = \frac{10!}{3!(10-3)!} = 120 \)
hockey-stick identity
The hockey-stick identity is a delightful result in combinatorics related to binomial coefficients. It visually resembles a hockey stick when plotted, which is how it gets its name. This identity provides a neat way to sum specific sequences of binomial coefficients.
Mathematically, the hockey-stick identity states that:
\[\sum_{r=0}^{m} \binom{n+r}{k} = \binom{n+m+1}{k+1} - \binom{n}{k+1}. \]
This is incredibly useful for computing the sum of combinations in a straightforward manner, simplifying calculations that might otherwise seem complex.
Mathematically, the hockey-stick identity states that:
\[\sum_{r=0}^{m} \binom{n+r}{k} = \binom{n+m+1}{k+1} - \binom{n}{k+1}. \]
This is incredibly useful for computing the sum of combinations in a straightforward manner, simplifying calculations that might otherwise seem complex.
- Imagine a vertical column of binomial coefficients starting at \( \binom{n}{k+1} \).
- Then there is a diagonal running from \( \binom{n}{k+1} \) upwards to \( \binom{n+m}{k+1} \).
summation of combinations
Summation of combinations is a technique in combinatorics used to handle problems where you need to add together a series of binomial coefficients. This approach is very useful for breaking down complex combinatorial problems into manageable parts.
In essence, if you have a sum like \( \sum_{r=0}^{m} \binom{n+r}{k} \), applying identities like the hockey-stick identity can simplify the calculation. The benefit of summing combinations is that it converts what might appear as a complicated problem into a simpler arithmetic one, leveraging known patterns and shortcuts.
In essence, if you have a sum like \( \sum_{r=0}^{m} \binom{n+r}{k} \), applying identities like the hockey-stick identity can simplify the calculation. The benefit of summing combinations is that it converts what might appear as a complicated problem into a simpler arithmetic one, leveraging known patterns and shortcuts.
- It allows for the simplification of nested or repeated summations.
- Helps in identifying patterns or relationships between different combinations.
Other exercises in this chapter
Problem 45
In a certain test there are \(n\) questions. In this test \(2^{k}\) students gave wrong answers to at least \((n-k)\) questions, where \(k=0,1,2, \ldots, n\). I
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The number of permutations of the letters \(a, b, c, d\) such that \(b\) does not follow \(a, c\) does not follow \(b\), and \(d\) does not follow \(c\), is (A)
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The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of a man and a woman, is (A) 1240 (B) 1840 (C) 1820 (D) 2005
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Suman writes letters to his five friends. The number of ways can be letters be placed in the envelopes so that atleast two of them are in the wrong envelopes ar
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