Short answer (I write nPr = <em>P(n, r)</em> and nCr = <em>C(n, r)</em> ):
<em>P</em> (6, 4) = 6! / (6 - 4)! = 6! / 2! = 720 / 2 = 360
<em>C</em> (6, 4) = <em>P</em> (6, 4) / 4! = 6! / (4! (6 - 4)!) = 360 / 24 = 15
Long answer:
<em>P(n, r)</em> counts the number of <u>permutations</u> of <em>n</em> objects taken <em>r</em> at a time, given by
<em>P(n, r)</em> = <em>n </em>! / (<em>n</em> - <em>r </em>)!
A permutation is a unique arrangement of objects such that the order in which they are arranged is taken into account. For example, if the objects in question are the numbers in the set {1, 2, 3}, then
• there are 3! = 6 total possible permutations if we take all 3 numbers at once:
123, 132, 213, 231, 312, 321
• there are 3!/(3-2)! = 3!/1! = 6 total permutations if we only take 2 numbers at once:
12, 13, 21, 23, 31, 32
• there are 3!/(3-1)! = 3!/2! = 3 total permutations if we take only 1 number at a time:
1, 2, 3
• and there is 3!/(3-0)! = 3!/3! = 1 way of permuting the 3 numbers without taking any of them:
(the permutation itself is just empty space)
By contrast, <em>C(n, r)</em> counts the <u>combinations</u> of <em>n</em> items taken <em>r</em> at a time, given by
<em>C(n, r)</em> = <em>P(n, r)</em> / <em>r </em>!
A combination is like a permutation, but the order of the objects doesn't matter. Continuing with the previous example of arrangements of the numbers from {1, 2, 3}, we have
• 3! / (3! (3-3)!) = 1 combination taking all 3 numbers at once:
123
(the other 5 permutations listed earlier are made up of the same numbers, so we consider them duplicates)
• 3! / (2! (3-2)!) = 3 combinations taking only 2 numbers at once:
12, 13, 23
• 3! / (1! (3-1)!) = 3 combinations taking only 1 number:
1, 2, 3
• 3! / (0! (3-0)!) = 1 combination taking none of them:
(again, empty space)
The main point is that <u>the order of objects is considered across permutations, while it's ignored across combinations</u>.
