Question

You know how to play 21 songs on your guitar. (a) If you want to choose 8 of the songs to make a setlist to play for a small gathering, how many possible setlists can you make (order matters here)? (b) If you want to choose 4 of the songs to teach to your youngest sibling, how many possible sets of 4 songs are there (order does not matter here)

Answer 1

(a) 8,204,716,800

(b) 5,985

Explanation:

Combinations and Permutations

Combinatorics is the part of the discrete mathematics that studies the enumeration of groups or sorting of a determined number of elements. The concept of combinations is tied to the different forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in a different order.

The concept of combinations is tied to the differents forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in different order.

On the other hand, permutations or variations are sets selected in a specific order and another set with the same element but in different order is considered a different set.

If we have n elements available to pick from in sets of m elements each, there can be C(n,m) different combinations, and it's given by

`\displaystyle C(n,m)=(n!)/(m!(n-m)!)`

Similarly the number of permutations is given by

`\displaystyle P(n,m)=(n!)/((n-m)!)`

(a) I have n=21 songs to pick from and I want to choose m=8 of them where the order matters, so it's a permutation:

`\displaystyle P(21,8)=(21!)/((13)!)=(51,090,942,171,709,440,000)/(6,227,020,800)=8,204,716,800`

I can make more than 8 billion different setlists

(b) To choose m=4 songs from n=21 songs where the order does not matter, we compute the combination

`\displaystyle C(21,4)=(21!)/(4!(17)!)=(21\cdot 20\cdot 19\cdot 18\cdot 17!)/(4\cdot 3\cdot 2\cdot 1(17)!)`

`\displaystyle C(21,4)=(143,640)/(24)=5,985`

I can make almost 6,000 sets of 4 songs