Problem 127 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \({ }^{56} \mathrm{C}_{3}\) (D) \({ }^{56} C_{4}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Binomial Coefficient

The notation \( { }^n C_r \) represents the binomial coefficient, which is calculated as \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). This coefficient gives us the number of ways to choose \( r \) objects from \( n \) objects without regard to order.

Step 2: Separating the Expressions

The expression given is \( { }^{50} C_{4} + \sum_{r=1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3} \). This consists of two parts: the single binomial coefficient \( { }^{50} C_{4} \), and the summation \( \sum_{r=1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3} \). Our task is to compute both and then add their results.

Step 3: Evaluate the Summation Term

We need to calculate \( \sum_{r=1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3} \) which involves finding \( { }^{55} C_{3} + { }^{54} C_{3} + { }^{53} C_{3} + { }^{52} C_{3} + { }^{51} C_{3} + { }^{50} C_{3} \). This simplification is possible due to the properties of binomial coefficients and can be represented as \( { }^{56} C_{4} \) using the symmetry property \( { }^n C_r = { }^n C_{n-r} \).

Step 4: Add the Results

Notice \( { }^{50} C_{4} \) from the original expression is exactly one of the terms within \( \sum_{r=1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3} \) because it is equivalent to \( { }^{50} C_{4} \). This means once you've summed the terms in Step 3, effectively the original \( { }^{50} C_{4} \) is already counted, leading us directly to the identity: \( { }^{50} C_{4} + \, \sum_{r=1}^{6}{\underline{\phantom{xx}}}^{56-r} C_{3} = { }^{56} C_{4} \).

Step 5: Determine the Correct Answer

The equivalent expression we've computed for the full original statement is \( { }^{56} C_{4} \). Therefore, the correct choice for this exercise is option (D).

Key Concepts

Combinatorics

In the realm of mathematics, combinatorics is the study of counting, arrangement, and combination of objects. It's like figuring out how many different ways you can organize a small group from a larger set.
In this exercise, we focus on the concept of binomial coefficients, which are crucial in combinatorics. Binomial coefficients tell us how many different ways we can select a subset of items from a larger set without caring about the order of selection. For instance, selecting 3 fruits from a basket of 10.

• The notation used is \( ^nC_r \) which represents the binomial coefficient.
• It's computed using the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n! \) represents the factorial of \( n \).

Combinatorics makes complex counting simple, breaking it down into understandable steps. So when you see a problem like choosing 4 members out of 50, you're stepping into the combinatorics field.

Symmetry Property

The symmetry property in binomial coefficients is a fascinating and useful insight. It states that the number of ways to choose \( r \) objects from \( n \) is the same as the number of ways to choose \( n-r \) objects from \( n \).
This idea can be represented mathematically as \( ^nC_r = ^nC_{n-r} \). This means if you're choosing 3 items out of 10, it's equivalent to choosing 7 items out of 10.

• It helps simplify computations involving series of binomial coefficients.
• The application of symmetry is effectively demonstrated in the expression \( \sum_{r=1}^{6} ~ ^{56-r}C_{3} \).
• Understanding symmetry allows you to manipulate expressions more efficiently.

Leveraging this property can reveal simpler, unexpected results in combinatorial calculations.

Summation of Binomial Coefficients

Summing binomial coefficients helps in various combinatorial problems. In this exercise, we observed how a series of binomial coefficients can be simplified into a single, elegant expression.
The problem contains an expression like \( \sum_{r=1}^{6} ~ ^{56-r}C_{3} \), which might look complicated initially, but properties like symmetry help simplify it. Here are some key points:

• The summation involves coefficients such as \( ^{55}C_3, ^{54}C_3, ... \).
• Thanks to binomial properties, these terms rearrange neatly to form a recognized pattern.
• The result adapts into \( ^{56}C_4 \), a transformation that's direct yet powerful.

Learning to break down and sum series of binomial coefficients is crucial in solving complex mathematical problems. This technique turns seemingly daunting tasks into manageable calculations.