From a pool of six juniors and eight seniors three co captains will be chosen for the football team. How many different combinations are possible if one junior and two seniors are chosen?

6c1 * 8c2

= 6 x 8 x 7/2
= 168

To find the number of different combinations possible, we can use the concept of combinations.

First, we need to choose one junior from a pool of six juniors. This can be done in 6 ways.

Next, we need to choose two seniors from a pool of eight seniors. The number of ways to choose two seniors from a pool of eight can be calculated using the combination formula:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items to choose from, and r is the number of items to choose.

In this case, n = 8 (seniors) and r = 2, so the number of ways to choose two seniors from a pool of eight is:

C(8, 2) = 8! / (2! * (8-2)!) = 8! / (2! * 6!) = (8 * 7) / (2 * 1) = 28

Finally, we need to multiply the number of ways to choose one junior by the number of ways to choose two seniors:

6 * 28 = 168

Therefore, there are 168 different combinations possible if one junior and two seniors are chosen to be co-captains of the football team.

To determine the number of different combinations of choosing one junior and two seniors from a pool of six juniors and eight seniors, we can use the concept of combination.

The formula for combination is represented as:
nCr = n! / (r!(n-r)!)

Where n is the total number of items to choose from and r is the number of items to be selected.

In this case, we have six juniors and eight seniors. We need to select one junior and two seniors. Therefore, n = 6 + 8 = 14 (total number of items) and r = 1 (number of juniors) + 2 (number of seniors) = 3.

The number of different combinations can be calculated as follows:

14C3 = 14! / (3!(14-3)!)
= 14! / (3! * 11!)

To determine the value of 14!, we multiply all the integers from 1 to 14. Similarly, to determine the value of 3!, we multiply all the integers from 1 to 3, and to determine the value of 11!, we multiply all the integers from 1 to 11.

Calculating the factorials:

14! = 14 * 13 * 12 * 11!
3! = 3 * 2 * 1
11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Now, we can substitute these values back into the combination formula:

14C3 = (14 * 13 * 12 * 11!) / (3 * 2 * 1 * (11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1))

Calculating the values:

14C3 = (14 * 13 * 12) / (3 * 2 * 1)

14C3 = (2184) / (6)

14C3 = 364

Therefore, there are 364 different combinations possible if one junior and two seniors are chosen.