Question

Are these permutation or combination problems?

Answer 1

Answer:

1. 63,960 total permutations

2. 24 or 64

Step-by-step explanation:

They are permutation problems, because the order of the elements matter (because 1-2-3 is different than 3-2-1).

In a combination, the order of the elements don't matter (so, 1-2-3 is the same as 3-2-1).

1. Lock combination.

You can choose from 41 different numbers and you have to pick 3. The formula is:

$P(n,r) = \frac{n!}{(n-r)!} = \frac{41!}{(41-3)!} = 63960$

Where n is the total numbers possible (41) and r is the quantity of numbers you have to pick (3).

For the first number, you can choose from 41 different numbers, so there are 41 possibilities.

For the second number, you cannot choose the same as the first one, so you are now limited to 40 possibilities.

For the third number, you are down to 39 possibilities since you cannot pick the same as the first two ones.

So, you have 41 * 40 * 39 = 63,960 possibilities in total.

2. Construct a 3-digit number from digits 4,6,8,9

This question doesn't specify if you can pick the same digit twice or not.

If you cannot, then the same formula as above applies:

$P(n,r) = \frac{n!}{(n-r)!} = \frac{4!}{(4-3)!} = 24$

Where n is the total numbers possible (4) and r is the quantity of numbers you have to pick (3).

If you can choose the same digit twice, then the formula is:

$n^{r} = 4^{3} = 64$

64 different numbers if you can pick the same number twice... which is logic since for the first number you have a choice among 4, same for the second and third number... so 4 * 4 * 4 = 64.