Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \({ }^{50} C_{4}+\sum_{r-1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3}\) is (A) \({ }^{55} C_{4}\) (B) \({ }^{55} C_{3}\) (C) \({ }^{56} C_{3}\) (D) \({ }^{56} C_{4}\)

Step-by-Step Solution

Answer

The correct choice is (D) \(^{56}C_{4}\).

Step 1: Understanding the Expression

The expression given is \[^{50} C_{4} + \sum_{r=1}^{6} { }^{56-r} C_{3}.\] We need to find the equivalent combinatorial expression from the answer choices.

Step 2: Simplifying the Summation

The summation part is \[\sum_{r=1}^{6} { }^{56-r} C_{3},\] which means we need to add the combinations for each \(r\) from 1 to 6: \[{ }^{55} C_{3} + { }^{54} C_{3} + { }^{53} C_{3} + { }^{52} C_{3} + { }^{51} C_{3} + { }^{50} C_{3}.\]

Step 3: Identifying the Pattern

Notice the pattern: \[^{55}C_{3} + { }^{54}C_{3} + { }^{53}C_{3} + { }^{52}C_{3} + { }^{51}C_{3} + { }^{50}C_{3}\]resembles \[^{56}C_{4}.\] This is because the identity in combinatorics states \[^{n}C_{r} + { }^{n-1}C_{r} + \dots = { }^{n+1}C_{r+1}.\] For \(n = 55\) and \(r = 3\), the identity confirms \[^{56}C_{4}\] as the result.

Step 4: Combining Parts of the Expression

Now combine both parts of the expression: \[^{50} C_{4} + { }^{56}C_{4}.\] According to Pascal's identity, \[^{n}C_{k} + { }^{n}C_{k+1} = { }^{n+1}C_{k+1}.\]Thus, combining gives \[^{56}C_{4},\] which matches with answer choice (D).

Key Concepts

Binomial Coefficients

Binomial coefficients are a fundamental concept in combinatorics that help us understand the number of ways to choose a subset of items from a larger set. They are represented mathematically as \(^{n}C_{k}\), which reads as "n choose k." This formula gives us the number of ways to select \(k\) items from \(n\) items without considering the order. The formula to determine a binomial coefficient is given by:

\[^{n}C_{k} = \frac{n!}{k!(n-k)!}\]where \( n! \) (n factorial) is the product of all positive integers up to \( n \), and \( k! \) is the factorial of \( k \).

• Used in probability to calculate the likelihood of events occurring.
• Critical in binomial expansions, which are expressions of the form \((a+b)^n\).
• Shows symmetry, as \(^{n}C_{k} = { }^{n}C_{n-k}\).

Understanding binomial coefficients allows us to solve problems involving combinations efficiently.

Combinatorial Identities

Combinatorial identities are equations that describe relationships between different binomial coefficients. They are crucial for simplifying complex combinatorial expressions. One such fundamental identity in combinatorics is:

\[^{n}C_{r} + { }^{n-1}C_{r} + \ldots + { }^{k}C_{r} = { }^{n+1}C_{r+1}\]This identity allows us to simplify summations of consecutive binomial coefficients. In our problem, the expression \(^{55}C_{3} + { }^{54}C_{3} + \ldots + { }^{50}C_{3}\) simplifies to \(^{56}C_{4}\).

• Use these identities to subtract, add, or simplify binomial coefficients.
• These identities work under specific conditions, like fixed \( r \) values in summations.
• Recognizing patterns in binomial expansions leveraging these identities makes for faster problem-solving.

Mastering combinatorial identities is key for efficiently handling algebraic combinatorial problems.

Pascal's Identity

Pascal's identity is a famous identity in combinatorics that provides a way to compute binomial coefficients recursively. It is expressed as:

\[^{n}C_{k} = { }^{n-1}C_{k-1} + { }^{n-1}C_{k}\]This relationship tells us that any entry in a row of Pascal’s triangle (which is a triangular arrangement of numbers that represent binomial coefficients) is the sum of the two numbers directly above it. For example, if you have an expression \(^{n}C_{k} + { }^{n}C_{k+1}\), it results in \(^{n+1}C_{k+1}\).

• This identity helps in expanding binomial expressions.
• Useful for deriving other combinatorial identities.
• Helps visualize binomial coefficients using Pascal's triangle.

Pascal's identity is a simple yet powerful tool in combinatorics, providing methods to understand and solve binomial expressions efficiently.