Question

There is a class or 20 pupils. The teacher decides to select any 6 pupils form them do attempt the hardest project on his list. How many different combinations of pupils may he choose?

Answer 1

`38760`

Explanation:

This is; given 20 choose 6.

The actual word 'combinations' is used in the question.

Combination is when the order does not matter `->color(white)()^nC_r` That is, in this context both the order `ab` and the order `ba` are just a count of 1 as `a` combined with `b` exist. Consequently this has a lower count than that of permutations where the order does matter.

Permutations has a higher count as order does matter `->color(white)()^nP_r" "`That is, in this context the order `ab` is counted and also the order `ba` is counted as an additional 1 giving a total of 2.

Note that `color(white)()^nP_r->(n!)/((n-r)!)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using an example as explanation `4!->4xx3xx2xx1`

Using the same example; factorial is sometimes shown as `" "ul(4|)`

We are using:

`" "color(white)()^nC_r = (n!)/((n-r)!r!)`

where `n=20 and r=6` giving

`" "(20!)/((20-6)!6!)`

Note that for calculation purposes you do not need to include `xx1` but I have done so just to show the completeness of the process.

In factorials it is always worth while looking for values that cancel. It does not take much to end up with very big numbers.

`" "(20xx19xx18xx17xx16xx15xxcancel(14!) )/(cancel(14!)xx6xx5xx4xx3xx2xx1)`
`color(white)()`

`" "20/(5xx4)xx19xx18/6xx17xx16/2xx15/3xx1`
`color(white)()`

`" "1xx19xx3xx17xx8xx5xx1 = 38760`