Combination Formula Lottery

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How many combinations of 3 numbers are possible from 10 numbers?

How many combinations of 3 numbers are possible from 10 numbers?

I am assuming it isn’t a random draw like rolling a ten-sided die, in which the same number can turn up twice or even three times, but it is a “no number can be chosen a second time” situation, i.e. if I choose a number a, it is simply not available as a choice when I choose again.

Pulling a lottery ball out of a bag or a tumble-them-up machine spitting one out and it is then put into a display means that the ball is not in the bag or the machine for subsequent drawings and the situation is as I specify it should be.

Your headline question: There are 10 different ways of choosing your first number, 9 different ways of then choosing your second number and 8 different ways of then choosing your third number, so:

10 x 9 x 8 = no of ways of choosing any 3 from 10 = 720

You might want to divide that by 6 (= 120) if order doesn’t matter, i.e. if you regard a, b, c as the same as a, c, b and b. c, a and b, a, c and c, a, b, and c, b, a. But if you regard these six outcomes (same 3 numbers but in a different order) as different from one another, then it is 720.

(1) To choose any 7 from 10 would mean 10 x 9 x 8 x 7 x 6 x 5 x 4 different outcomes = 604,800 but any of these outcomes can be chosen in 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040 different orders, so the answers are 604,800 if order matters and 604,800 / 5,040 = 120 if the order doesn’t matter.

(2) To choose any 6 from 10 would mean 10 x 9 x 8 x 7 x 6 x 5 different outcomes = 151,200 but any of these outcomes can be chosen in 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders, so the answers are 151,200 if order matters and 151,200/ 720 = 210 if the order doesn’t matter.

(3) To choose any 5 from 10 would mean 10 x 9 x 8 x 7 x 6 different outcomes = 30,240 but any of these outcomes can be chosen in 5 x 4 x 3 x 2 x 1 = 120 different orders, so the answers are 151,200 if order matters and 30,240/ 120 = 252 if the order doesn’t matter.

(4) To choose any 4 from 10 would mean 10 x 9 x 8 x 7 different outcomes = 5,040 but any of these outcomes can be chosen in 4 x 3 x 2 x 1 = 24 different orders, so the answers are 5.040 if order matters and 5.040/ 24 = 210 if the order doesn’t matter.

(5) To choose any 3 from 10 has already been demonstrated to be 720 if order does matter and 120 if it doesn’t matter.

It is a simple matter to generalise this approach to define the formula you seek, which is: n! / (n-r)! x r! where n is the total number of lottery balls of which r are to be chosen.

[Anyone familiar with Pascal's Triangle will recognise the numbers 120, 210 and 252 as coefficients of (a+b)^10]

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