A committec of five teachers is to be chosen from six French teachers and six Music teacher In how many ways can this be done so that the committee will consist of: (a) four French teachers; (b) at least one French teacher, (c) at most two Music teachers?

To solve these problems, we will use combinatorics. (a) To form a committee of four French teachers from six French teachers, we need to choose all members from this group. The number of ways to choose 4 French teachers from 6 is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items to choose from, and k is the number of items to choose. So, for four French teachers: C(6, 4) = 6! / (4! * 2!) = 15 ways. We also need to choose one Music teacher from six available Music teachers: C(6, 1) = 6 ways. Thus, the total number of ways to form this committee is: 15 * 6 = 90 ways. (b) To find the number of committees with at least one French teacher, we can first calculate the total number of committees without restrictions and then subtract the number of committees with no French teachers. Total ways to choose any five teachers from twelve (6 French + 6 Music): C(12, 5) = 12! / (5! * 7!) = 792 ways. Now, we will calculate the number of committees with no French teachers, which means all selected teachers must be Music teachers. To form a committee of five Music teachers from six, we get: C(6, 5) = 6 ways. Now, we subtract the committees with no French teachers from the total: 792 - 6 = 786 ways. (c) To find the number of committees consisting of at most two Music teachers, we can consider the cases of 0, 1, and 2 Music teachers. - **Case 1**: 0 Music teachers (all 5 are French): C(6, 5) = 6 ways. - **Case 2**: 1 Music teacher (4 French teachers): C(6, 4) * C(6, 1) = 15 * 6 = 90 ways. - **Case 3**: 2 Music teachers (3 French teachers): C(6, 3) * C(6, 2) = 20 * 15 = 300 ways. Now, we add the ways from all cases together: 6 + 90 + 300 = 396 ways. Thus, the results are: (a) 90 ways (b) 786 ways (c) 396 ways