A group of 25 lecturers includes six mathematicians. in how many ways can a committee of 4 be selected from this group to include at least one mathematician?
6 * 19P3
Oops. I meant 6*19C3
To find the number of ways to select a committee of 4 from a group of 25 lecturers, including at least one mathematician, you can use the principle of inclusion-exclusion.
Step 1: Find the total number of ways to select a committee of 4 from a group of 25 lecturers, regardless of whether they are mathematicians or not. This can be calculated using the combination formula, given by:
C(n, r) = n! / (r!(n-r)!),
where n is the total number of lecturers and r is the desired committee size.
In this case, the total number of lecturers is 25 and the committee size is 4. So, the total number of ways to select a committee of 4 from 25 lecturers is:
C(25, 4) = 25! / (4!(25-4)!) = 12,650.
Step 2: Find the number of ways to select a committee of 4 from the group of 25 lecturers without considering the mathematicians at all.
Since there are 25 lecturers in total and 6 mathematicians, there must be 19 non-mathematicians. Therefore, the number of ways to select a committee of 4 non-mathematicians from the remaining 19 lecturers (excluding the mathematicians) is:
C(19, 4) = 19! / (4!(19-4)!) = 3,876.
Step 3: Find the number of ways to select a committee of 4 from the group of 25 lecturers containing only mathematicians.
Since there are 6 mathematicians in total, the number of ways to select a committee of 4 mathematicians from the 6 mathematicians is:
C(6, 4) = 6! / (4!(6-4)!) = 15.
Step 4: Find the number of ways to select a committee of 4 from the group of 25 lecturers that includes at least one mathematician by subtracting the number of committees without mathematicians from the total number of committees. This can be calculated as:
Total ways to select committee with at least one mathematician = Total ways to select committee - Ways to select committee with no mathematicians
= 12,650 - (3,876 + 15)
= 12,650 - 3,891
= 8,759.
Therefore, there are 8,759 ways to select a committee of 4 from a group of 25 lecturers to include at least one mathematician.