Question

In the expansion of \((x-y)^{10}\), (co-efficient of \(x^{7} y^{3}\) ) \(\left(\right.\) co-efficient of \(\left.x^{3} y^{7}\right)=\) (a) \({ }^{10} \mathrm{C}_{7}\) (b) \({ }^{2.10} \mathrm{C}_{7}\) (c) \({ }^{10} \mathrm{C}_{7}+{ }^{10} \mathrm{C}_{1}\) (d) 0

Short answer

None of the given options are correct. The expression (co-efficient of \(x^{7} y^{3}\) ) (co-efficient of \(x^{3} y^{7}\)) is equal to 120 * 120, and none of the options match this result.

Step-by-step solution

Write down the binomial theorem formula

The binomial theorem formula for the expansion of \((a+b)^n\) is given by: \((a+b)^n = \sum_{k=0}^n {n \choose k} a^{n-k} b^k\) In our case, the binomial expansion will look like: \((x-y)^{10} = \sum_{k=0}^{10} {10 \choose k} x^{10-k} (-y)^k\)

Find the coefficients of \(x^7y^3\) and \(x^3y^7\)

For the term with \(x^7y^3\), the value of k is 3. So, we will plug this value in the combination formula as mentioned earlier: \({10 \choose 3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}\) For the term with \(x^3y^7\), the value of k is 7. So, we will plug this value in the combination formula: \({10 \choose 7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!}\)

Solve for the coefficients

Now, we will solve the two combinations we found in Step 2: \({10 \choose 3} = \frac{10!}{3!7!} = \frac{10\times9\times8\times7!}{3\times2\times1\times7!} = \frac{10\times9\times8}{6} = 120\) \({10 \choose 7} = \frac{10!}{7!3!} = \frac{10\times9\times8\times7!}{7!3\times2\times1} = \frac{10\times9\times8}{6} = 120\)

Check the answer options and conclude the solution

The coefficients of \(x^7y^3\) and \(x^3y^7\) are equal: 120 = 120 So, the expression (co-efficient of \(x^{7} y^{3}\) ) (co-efficient of \(x^{3} y^{7}\)) = 120 * 120 Now let's check the given options: (a) \({ }^{10} \mathrm{C}_{7}\) = 120. This option represents only the coefficient of one term, so it's not the correct answer. (b) \({ }^{2.10} \mathrm{C}_{7}\) = This option doesn't make sense mathematically. (c) \({ }^{10} \mathrm{C}_{7}+{ }^{10} \mathrm{C}_{1}\) = 120 + \({ }^{10}C_1\) = 120 + 10 = 130. This option represents the sum of the coefficients, not the expression we want. (d) 0: This is not the correct answer since the coefficients are not zero. None of the given options match the correct expression which is the product of the two coefficients. There is no correct answer among the given options.

Key concepts

Coefficient Calculation

Understanding coefficient calculation is essential when dealing with the Binomial Theorem. The coefficient of any term in the binomial expansion can be determined using combinatorial mathematics. In the expression \((x-y)^{10} = \sum_{k=0}^{10} {10 \choose k} x^{10-k} (-y)^k\), the formula \({n \choose k} = \frac{n!}{k!(n-k)!} \) is used to find the coefficients.
Here, \(n\) is the total number of trials or choices, and \(k\) is the number of times an outcome is chosen. For example, to calculate the coefficient of \(x^7y^3\), you choose \(k=3\) and compute:

• \({10 \choose 3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)

Similarly, for \(x^3y^7\), choose \(k=7\):

• \({10 \choose 7} = \frac{10 imes 9 imes 8}{3 imes 2 imes 1} = 120\)

Knowing how to calculate these coefficients precisely explains why both terms \(x^7y^3\) and \(x^3y^7\) have the same coefficient. These calculations are foundational for gaining a mastery over polynomial expansions and expression analysis.

Combinatorics

Combinatorics is the mathematical study of counting and arranging possibilities, and it plays a vital role in the Binomial Theorem. When calculating the coefficients in a binomial expansion, combinatorial methods, like combinations, are primarily used.
Combination, or choice, in mathematics is represented by the binomial coefficient. It's the number of ways to choose \(k\) items from a total of \(n\), written as \({n \choose k}\).
The formula for combinations is also a great example of how elements can be repeatedly selected without considering the order:

• \({n \choose k} = \frac{n!}{k!(n-k)!}\)

This expression is found in the broader binomial formula\( (a+b)^n = \sum_{k=0}^n {n \choose k} a^{n-k} b^k \). In our exercise, identifying the correct values of \(k\) helps determine the terms \(x^7y^3\) and \(x^3y^7\).
By understanding combinatorics' role in the Binomial Theorem, it becomes significantly easier to expand and analyze polynomials accurately, laying a crucial foundation for advanced algebra and probability theory.

Polynomial Expansion

Polynomial Expansion often involves using the Binomial Theorem to simplify expressions like \( (x-y)^{10} \). The theorem allows us to expand a binomial raised to any power in a systematic way. The general iziation \( (a+b)^n = \sum_{k=0}^n {n \choose k} a^{n-k} b^k \)provides the structure for expansion.
Each term in the expansion is a product of combinations and powers of the variables. For instance, in our specific exercise of \((x-y)^{10}\), it helps us find each term, such as \(x^7y^3\), by calculating \({10 \choose 3} x^{7}(-y)^3\).
The negative sign in \(-y\) signifies that each iteration requires assessing both coefficients and signs. By breaking complicated expressions into straightforward products, the Binomial Theorem simplifies polynomial expansion, making it manageable and precise.
Understanding this concept is vital for tackling complex algebraic expressions, especially in calculus and various fields of applied mathematics, ensuring a well-rounded grasp of polynomial behavior and transformations.