Why is 6P4 = 360 but 6C4 = 15?

Short answer (I write nPr = P(n, r) and nCr = C(n, r) ):

P (6, 4) = 6! / (6 - 4)! = 6! / 2! = 720 / 2 = 360

C (6, 4) = P (6, 4) / 4! = 6! / (4! (6 - 4)!) = 360 / 24 = 15

Long

P(n, r) counts the number of permutations of n objects taken r at a time, given by

P(n, r) = n ! / (n - r )!

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, C(n, r) counts the combinations of n items taken r at a time, given by

C(n, r) = P(n, r) / r !

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 the order of objects is considered across permutations, while it's ignored across combinations.

70,000 + 6,000 + 800 + 30

step-by-step explanation:

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step-by-step explanation: