This is a classic combinatorics problem involving combinations. The formula for combinations is C(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, n = 10 and r = 4, so C(10, 4) = 10! / (10-4)! = 10! / 6! = 210.

To solve this probability problem, we can use conditional probability. First, we calculate the probability of selecting 2 red balls from the 6 red balls: P(2 red) = C(6, 2) / C(12, 2) = 15 / 66. Then, given that we have selected 2 red balls, we calculate the probability of selecting 1 blue ball from the remaining 10 balls: P(1 blue | 2 red) = C(4, 1) / C(10, 1) = 4 / 10. Multiplying these probabilities, we get P(2 red and 1 blue) = P(2 red) * P(1 blue | 2 red) = (15 / 66) * (4 / 10) = 1/10.

This question relates to the concept of maximum matching in graph theory. According to Tutte's Theorem, the maximum number of edges that can be added to a graph without creating a cycle is equal to the minimum number of vertices that need to be removed to make the graph acyclic. In this case, we can use Tutte's Theorem to find that the maximum number of edges that can be added is 3.

The chromatic number of a graph is the minimum number of colors needed to color the vertices of the graph such that no two adjacent vertices have the same color. In this case, the graph has 4 vertices and 6 edges, which means it is a complete graph K_4. The chromatic number of K_4 is known to be 4, so the correct answer is 4.

A tree is a connected graph with no cycles. A complete graph with n vertices has n(n-1)/2 edges. To make a complete graph a tree, we need to remove n-1 edges while maintaining connectivity. Therefore, the minimum number of edges to be removed from a complete graph with 10 vertices to make it a tree is 10-1 = 9.

This is a classic combinatorics problem involving permutations. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, n = 15 and r = 3, so P(15, 3) = 15! / (15-3)! = 15! / 12! = 2205.

To solve this problem, we can use the principle of inclusion-exclusion. First, we calculate the total number of ways to select a team of 3 employees from 10 employees: C(10, 3) = 10! / (10-3)! = 120. Then, we calculate the number of ways to select a team of 3 employees with no women: C(5, 0) * C(5, 3) = 1 * 10 = 10. Finally, we subtract the number of ways to select a team with no women from the total number of ways to select a team to get the number of ways to select a team with at least 1 woman: 120 - 10 = 110.

This is a classic probability problem involving binomial distribution. The formula for binomial distribution is P(x) = (n! / (x!(n-x)!)) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 5, p = 1/2, and q = 1/2. Plugging these values into the formula, we get P(5) = (10! / (5!(10-5)!)) * (1/2)^5 * (1/2)^(10-5) = 252 / 1024 = 1/4.

This is a classic combinatorics problem involving combinations. The formula for combinations is C(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, we need to select 5 people from a group of 20, but 2 specific people must be included. So, we first select the 2 specific people, which can be done in C(2, 2) = 1 way. Then, we select the remaining 3 people from the remaining 18 people, which can be done in C(18, 3) = 816 ways. Multiplying these values, we get the total number of ways to select a committee of 5 people with 2 specific people included: 1 * 816 = 816.

This question relates to the concept of maximum matching in graph theory. According to Tutte's Theorem, the maximum number of edges that can be added to a graph without creating a cycle is equal to the minimum number of vertices that need to be removed to make the graph acyclic. In this case, we can use Tutte's Theorem to find that the maximum number of edges that can be added is 2.

The chromatic number of a graph is the minimum number of colors needed to color the vertices of the graph such that no two adjacent vertices have the same color. In this case, the graph has 6 vertices and 9 edges, which means it is a complete graph K_6. The chromatic number of K_6 is known to be 4, so the correct answer is 4.

A tree is a connected graph with no cycles. A complete graph with n vertices has n(n-1)/2 edges. To make a complete graph a tree, we need to remove n-1 edges while maintaining connectivity. Therefore, the minimum number of edges to be removed from a complete graph with 8 vertices to make it a tree is 8-1 = 7.

This is a classic combinatorics problem involving permutations. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected. In this case, n = 12 and r = 3, so P(12, 3) = 12! / (12-3)! = 12! / 9! = 1320.

To solve this problem, we can use the principle of inclusion-exclusion. First, we calculate the total number of ways to select a team of 4 employees from 15 employees: C(15, 4) = 15! / (15-4)! = 15! / 11! = 1365. Then, we calculate the number of ways to select a team of 4 employees with no women: C(9, 0) * C(6, 4) = 1 * 15 = 15. Finally, we subtract the number of ways to select a team with no women from the total number of ways to select a team to get the number of ways to select a team with at least 2 women: 1365 - 15 = 1350.