Question

6 keys are to be placed on a keyring. What is the number of ways of arranging the keys if there are 3 identical locker keys and 2 identical car keys?

Answer 1

10 ways

Explanation:

First off, this is a permutation problem because we care about the order of the keys. The general formula is:

`P_(n,k)=(n!)/((n-k)!); n="population", k="picks"`

If we were to arrange the 6 keys in a row and the keys were all different, we'd have:

`6! = 720` ways

But we have a few things we need to adjust for.

First off, we're arranging the keys in a circle. When arranging in a circle , there is no "first position" or "last position" like you get in a row. And so arrangements 1, 2, 3, 4, 5, 6 and 2, 3, 4, 5, 6, 1 are actually the same. It turns out we need to divide by the number of items in the ring :

`(6!)/6=(6xx5!)/6=5!`

The other thing we need to adjust for is the presence of identical keys . There are 3 identical locker keys and 2 identical car keys. And so if keys 1, 2, 3 are locker keys, arrangement 1, 2, 3, 4, 5, 6 is the same as 3, 2, 1, 4, 5, 6. And so on. To get rid of these duplicates, we divide by the number of ways we can arrange the duplicate keys, and that's `3! = 6` for the locker keys and `2! =2` for the car keys:

`(5!)/(3!2!)=120/(6xx2)=10`