- Utilize the binomial coefficient formula \({}_nC_k = \frac{n!}{k!(n-k)!}\) for \(n=6\).
- Systematically calculate each coefficient from \({}_6C_0\) to \({}_6C_6\):
- \({}_6C_0=1, {}_6C_1=6, {}_6C_2=15, {}_6C_3=20, {}_6C_4=15, {}_6C_5=6, {}_6C_6=1\).
- Match the resulting sequence \(1, 6, 15, 20, 15, 6, 1\) to the provided options.
- The correct choice is \(\boxed{b}\).
Explanation
- Understanding the Binomial Theorem
The binomial theorem states that the expansion of \((a+b)^n\) is given by the sum:
\[(a+b)^n = \sum_{k=0}^{n} {}_nC_k a^{n-k} b^k\]
where \({}_nC_k\) are the binomial coefficients calculated using the formula:
\[{}_nC_k = \frac{n!}{k!(n-k)!}\]
In our case, we are given \(n = 6\), so we need to determine the coefficients \({}_6C_0, {}_6C_1, {}_6C_2, {}_6C_3, {}_6C_4, {}_6C_5,\) and \({}_6C_6.\)
- Calculating the Binomial Coefficients
We calculate each coefficient using the formula \({}_6C_k = \frac{6!}{k!(6-k)!}\):
- \({}_6C_0 = \frac{6!}{0!6!} = 1\)
- \({}_6C_1 = \frac{6!}{1!5!} = \frac{6}{1} = 6\)
- \({}_6C_2 = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15\)
- \({}_6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\)
- \({}_6C_4 = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15\) (due to symmetry \({}_nC_k = {}_nC_{n-k}\))
- \({}_6C_5 = \frac{6!}{5!1!} = \frac{6}{1} = 6\)
- \({}_6C_6 = \frac{6!}{6!0!} = 1\)
- Conclusion and Final Selection
The sequence of binomial coefficients for \((a+b)^6\) is \(1, 6, 15, 20, 15, 6, 1\). Comparing this result with the given options:
A\(1, 6, 10, 15, 10, 6, 1\)
B\(1, 6, 15, 20, 15, 6, 1\)
C\(1, 6, 8, 10, 8, 6, 1\)
D\(1, 6, 20, 30, 20, 6, 1\)
It is clear that the correct set of coefficients matches option b.
Examples
Binomial coefficients are used in probability theory, such as determining the number of ways to pick a committee of a certain size from a larger group of people. For instance, if you have a class of 6 students and want to know how many ways you can form a group of 3, you use the coefficient \({}_6C_3\), which we found to be 20. This allows for efficient planning and resource allocation in various real-world scenarios.