Problem 105
Question
For any positive integers \(m, n\) (with \(n \geq m\) ), let and \(\left(\begin{array}{l}n \\ m\end{array}\right)={ }^{n} C_{m}\), then \(\left(\begin{array}{l}n \\ m\end{array}\right)+\left(\begin{array}{l}n-1 \\\ m\end{array}\right)+\left(\begin{array}{l}n-2 \\\ m\end{array}\right)+\ldots+\left(\begin{array}{c}m \\ m\end{array}\right)=\) (A) \(\left(\begin{array}{l}n+1 \\ m\end{array}\right)\) (B) \(\left(\begin{array}{l}n+1 \\ m+1\end{array}\right)\) (C) \(\left(\begin{array}{c}n \\ m+1\end{array}\right)\) (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (B) \( \binom{n+1}{m+1} \).
1Step 1: Understanding the Sum
The expression \( \binom{n}{m} + \binom{n-1}{m} + \binom{n-2}{m} + \cdots + \binom{m}{m} \) can be seen as a sum of combinations where the upper part of the binomial coefficient decreases while the lower part remains constant \( m \). This is essentially counting the number of ways to select items when varying the total number of available items.
2Step 2: Recognizing the Pattern
Notice that the pattern formed by this sum is similar to the binomial expansion terms. The sum \( \sum_{k=m}^{n} \binom{k}{m} \) represents selecting \( m \) items from at most \( n \) items.
3Step 3: Applying Identity
Using a combinatorial identity, we know that \( \sum_{k=m}^{n} \binom{k}{m} = \binom{n+1}{m+1} \). This identity checks all possible ways to choose \( m \) from sequences up to \( n \) plus another element, effectively considering an extra step not in the original range.
4Step 4: Verifying Options
Now, compare \( \binom{n+1}{m+1} \) with the given options. (A) \( \binom{n+1}{m} \)(B) \( \binom{n+1}{m+1} \)(C) \( \binom{n}{m+1} \)(D) None of these. The identity matches precisely with option (B), \( \binom{n+1}{m+1} \).
Key Concepts
Combinatorial IdentityBinomial Coefficient SumBinomial Expansion
Combinatorial Identity
Combinatorial identities play a crucial role in solving problems related to combinations and permutations. These identities allow us to simplify complex expressions involving binomial coefficients and transform them into easier-to-evaluate forms. One such identity is the shifting identity applied in the exercise given:
Recognizing and utilizing such identities is essential for working efficiently with combinatorial problems.
- It states that the sum of binomial coefficients from a fixed lower index up to an upper index, i.e., \( \sum_{k=m}^{n} \binom{k}{m} \), can be expressed as \( \binom{n+1}{m+1} \) .
Recognizing and utilizing such identities is essential for working efficiently with combinatorial problems.
Binomial Coefficient Sum
The summation of binomial coefficients involves computing the total number of ways to choose a specific number of items from a varying total. In the exercise, the sum \( \binom{n}{m} + \binom{n-1}{m} + \cdots + \binom{m}{m} \) is considered. Each term of this sum represents combinations where the lower term \( m \) remains constant while the upper term decreases.
- By summing these terms, we effectively account for every possible case where we can choose \( m \) elements from different total counts ranging from \( m \) to \( n \).
Binomial Expansion
Binomial expansion refers to expanding expressions of the form \((a+b)^n\) using binomial coefficients. Each expansion follows a fixed pattern rooted in Pascal's triangle, with each term taking the general form \( \binom{n}{k} a^{n-k}b^k \). This framework offers insights into how binomial coefficients behave under various algebraic manipulations.
- The exercise shows a conceptual similarity to binomial expansion. Although the exercise does not explicitly expand a binomial expression, the summation pattern is akin to selecting terms as it's done in binomial expansions.
Other exercises in this chapter
Problem 103
The number of words that can be formed, with the letters of the work 'Pataliputra' without changing the relative order of the vowels and consonants, is (A) 3600
View solution Problem 104
On a new year day every student of a class sends a card to every other student. The postman delivers 600 cards. The number of students in the class are (A) 42 (
View solution Problem 106
The number of 7 digit numbers the sum of whose digits is even, is (A) \(35 \times 10^{5}\) (B) \(45 \times 10^{5}\) (C) \(50 \times 10^{5}\) (D) None of these
View solution Problem 107
The number of ways of choosing \(m\) coupons out of an unlimited number of coupons bearing the letters \(A, B\) and \(C\) so that they cannot be used to spell t
View solution