The binomial coefficient (\binom{n}{k}) represents the number of ways to choose k items from a set of n distinct items, without regard to order. The formula (\binom{n}{k} = \frac{n!}{k!(n-k)!}) is derived from the fundamental principle of counting.

Using the formula (\binom{n}{k} = \frac{n!}{k!(n-k)!}), we can calculate (\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{120}{2\cdot6} = 10).

Pascal's Triangle is a triangular array of binomial coefficients, where each entry is the sum of the two entries above it. The first row of Pascal's Triangle consists of 1, the second row consists of 1 and 1, and so on.

Pascal's Identity states that the binomial coefficient (\binom{n}{k}) can be expressed as the sum of the two binomial coefficients (\binom{n-1}{k}) and (\binom{n-1}{k-1}). This identity is fundamental in the study of binomial coefficients and has various applications in combinatorics.

Using the binomial theorem, we can expand ((1+x)^{10}) as (\sum_{k=0}^{10} \binom{10}{k} x^k). The coefficient of (x^3) corresponds to (k=3), so we have (\binom{10}{3} = \frac{10!}{3!7!} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1} = 120).

To solve this problem, we can use the binomial coefficient (\binom{n}{k}). In this case, we have n = 6 (total number of people) and k = 3 (number of people in each group). So, the number of ways to divide 6 people into two groups of 3 people each is (\binom{6}{3} = \frac{6!}{3!3!} = \frac{6\cdot5\cdot4}{3\cdot2\cdot1} = 20).

Using the binomial theorem, we can expand ((2x-3)^{8}) as (\sum_{k=0}^{8} \binom{8}{k} (2x)^{8-k}(-3)^k). The coefficient of (x^5) corresponds to (k=3), so we have (\binom{8}{3} (2)^{8-3}(-3)^3 = \frac{8!}{3!5!} \cdot 2^5 \cdot (-3)^3 = 2720).

To solve this problem, we can use the binomial coefficient (\binom{n}{k}). In this case, we have n = 30 (total number of students) and k = 5 (number of students to be selected). So, the number of ways to select 5 students from a class of 30 students is (\binom{30}{5} = \frac{30!}{5!25!} = \frac{30\cdot29\cdot28\cdot27\cdot26}{5\cdot4\cdot3\cdot2\cdot1} = 142506).

Using the property (\binom{n}{k} = \binom{n}{n-k}), we can simplify (\binom{100}{99}) as (\binom{100}{100-99} = \binom{100}{1} = 100).

Since 2 specific people must be included in the committee, we need to select 2 more people from the remaining 10 people. So, the number of ways to form the committee is (\binom{10}{2} = \frac{10!}{2!8!} = \frac{10\cdot9}{2\cdot1} = 45). However, since the 2 specific people can be arranged in 2 different orders, we need to multiply this value by 2. Therefore, the total number of ways to form the committee is 2 (\cdot) 45 = 90.

Using the binomial theorem, we can expand ((x+2)^{12}) as (\sum_{k=0}^{12} \binom{12}{k} x^{12-k}2^k). The coefficient of (x^6) corresponds to (k=6), so we have (\binom{12}{6} 2^6 = \frac{12!}{6!6!} \cdot 2^6 = \frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1} \cdot 64 = 924).

Let (S) be the set of people who liked soccer, (B) be the set of people who liked basketball, and (S\cap B) be the set of people who liked both soccer and basketball. Then, by the principle of inclusion-exclusion, the number of people who liked only soccer or only basketball is (|S\cup B| = |S| + |B| - |S\cap B| = 60 + 40 - 20 = 80).

Using the binomial theorem, we can expand ((3x^2-2)^{6}) as (\sum_{k=0}^{6} \binom{6}{k} (3x^2)^{6-k}(-2)^k). The coefficient of (x^4) corresponds to (k=2), so we have (\binom{6}{2} (3x^2)^{6-2}(-2)^2 = \frac{6!}{2!4!} \cdot 3^4x^8 \cdot 4 = 1296x^8).

To solve this problem, we can use the multiplication principle. First, the club can select a president in 15 ways. Then, the club can select a vice president from the remaining 14 members in 14 ways. Finally, the club can select a secretary from the remaining 13 members in 13 ways. Therefore, the total number of ways to select a president, a vice president, and a secretary is 15 (\cdot) 14 (\cdot) 13 = 2730.