I need a mathematical formula?

Question

Specifically, hwo many sets of three can be made from a group of 26? What is the formula used to find this out?

Answer 1

C(26,3) = "26 choose 3"

The general formula for C(x, n) is:
x! / [ (x-n)! * n! ]

So you want:
26! / 23! * 3!

This is:
26 x 25 x 24 / 6

Or:
2600 ways

The easy way to think of it is to think of the number of ways to pick the first item: 26

Once you have picked the first item, there are 25 ways to pick the second item: 25

Finally there are 24 ways to pick the 3rd item.

That's 26 x 25 x 24.

However, sets don't care about order. The following sets are all equivalent {A, B, C}, {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, {C, B, A}. So you have to divide by the number of ways to arrange the 3 items in the same set. That's 3! or 6.

So the final answer is: 26 x 25 x 24 / 6 = 2600 ways to choose 3 items from a group of 26.

Answer 2

C(26,3)=
26!/(23!3!)=
(26*25*24)(1*2*3)=
2600

Answer 3

Pick one = 26 choices.
Pick another one = 25 choices.
Pick the third one = 24 choices.
Start again.

But after all possible picks, each set has come up six times, in a different order each time.

Therefore 26 * 25 * 24 / 6 = 2600 different sets of three.

Answer 4

You need to use combinations if order is unimportant, permutations if it is

Eg if the group of 26 were represented by all the letters of the alphabet and ABC is the same group as ACB ie order is not important, then

Number of groups = 26C3
= 26!/(23!3!)
= (26.25.24)/(3.2.1)
=2600

If order IS important ie ABC is different to ACB as in say a line of people, then

Number of groups = 26P3
= 26!/23!
= 26.25.24
= 15600

Note nCr = n!/[(n-r)!r!)
and nPr = n!/(n-r)!

where a! = a.(a-1).(a-2) .. 3.2.1

Answer 5

x= number of accounts I can call
I know: A - hours available
B = Hrs to complete a call
C= hours to make an incomplete call
65% of the time calls are completed