Basic combinatorics

Ronan5000

A lottery is run where balls numbered 1 to 30 are taken out of a bag. How many different
ways are there of choosing 3 numbers from these 30?

The answer given to this question from the book I'm using is 4060:
30x29x28 / 3x2x1 = 4060

However if you use factorials: 30! / (30-3)! you get 24360

Can anyone tell me which is the correct way and why?

Yakuza

4060 is correct.
You have 30 numbers to choose from for the first, 29 for the second and 28 for the 3rd. This means there are 30 * 29 * 28 ways to choose 3 numbers from 30.

However, of all those ways there are many repeated sequences (1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, 3 2 1) which amount to the same thing.
So, of the 3 numbers you've chosen, how many ways can they be arranged? 3! (3 choices for the first, 2 for the second and one for the third).

Therefore, you have to divide the larger number of permutations (full of repetition) by the number of ways to arrange the chosen numbers to come up with the number of unique combinations,

So, the answer is (30 * 29 * 28) / (3 * 2 * 1) = 4060.

More generally, this is n! / r! where n is the total number of objects and r the amount chosen.

If order doesn't matter, it's a combination (like in the Lotto draw).
If order does matter, it's a permutation. Strictly speaking, a combination lock should be called a permutation lock as order matters with them

Ficheall

Yakuza wrote: »

More generally, this is n! / r! where n is the total number of objects and r the amount chosen.

Just in case it throws anyone, you've a typo there - should be n!/(r!(n-r)!) .

Yakuza

Indeed, my bad.

Ronan5000

Dd