1. A group of 25 lecturers include six mathematician. In how many ways can a committee of 4 be selected from this group to include at least one mathematician?

2. In how many ways can the first, second and third positions be taken by ten candidates in a test, assuming that there must be no tie.

3. Six people sit at a round table. In how many ways can that be done if two of them must sit next to each other

#1

There are 6 ways to pick the mathematician. There are 19C3 ways to pick the others.

#2
There are 10P3 ways to pick 3 of 10 finishers.

#3
Consider the 2 people who must sit together as 1 unit. That means there are 5 individuals to seat around the table. So, there are just 4! ways to seat the people.

1. To solve this problem, we need to consider the different possibilities for selecting a committee of 4 members from a group of 25 lecturers with 6 mathematicians.

To determine the number of ways to select a committee with at least one mathematician, we need to subtract the number of ways to form a committee with no mathematicians from the total number of possible committees.

We can form a committee with no mathematicians by choosing 4 members from the remaining 19 lecturers who are not mathematicians. This can be done in the following way:

- Selecting 4 members out of 19: C(19, 4)

To find the total number of possible committees, we need to select 4 members out of the entire group of 25 lecturers. This can be done in the following way:

- Selecting 4 members out of 25: C(25, 4)

Therefore, the number of ways to select a committee of 4 members from a group of 25 lecturers to include at least one mathematician is:

C(25, 4) - C(19, 4)

2. To find the number of ways in which the first, second, and third positions can be taken by ten candidates in a test, assuming no tie is allowed, we need to consider the concept of permutations.

In the first position, any of the ten candidates can be chosen. After selecting the candidate for the first position, there are nine remaining candidates for the second position. Finally, after selecting candidates for the first two positions, there are eight remaining candidates for the third position.

The total number of ways to select candidates for each position is found by multiplying the number of options for each position:

10 * 9 * 8 = 720

Therefore, there are 720 possible ways for the first, second, and third positions to be taken by ten candidates in a test, assuming no tie is allowed.

3. To calculate the number of ways six people can sit at a round table if two of them must sit next to each other, we can treat the two individuals who must sit together as a single entity.

So, we now have five distinct entities (one group of two individuals and four remaining individuals) to arrange around a circular table.

The number of ways to arrange these entities around a circular table is given by (n - 1)!, where n is the number of entities. In this case, n = 5.

Therefore, the number of ways for six people to sit at a round table with two individuals sitting next to each other is (5 - 1)! = 4!.

4! = 4 x 3 x 2 x 1 = 24

Therefore, there are 24 possible ways for the six people to sit at a round table with two individuals sitting next to each other.