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

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: \((x-y)^{10} = \sum_{k=0}^{10} {10 \choose k} x^{10-k} (-y)^k\)

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: \({10 \choose 7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!}\)

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

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.