Variation number of tw0 digit numbers in array of 5 without repetition and order

Question

How to calculate variations of 2 digit numbers [00-99] in the array of 5 for each variation? Order and repetition should not be considered.

variation pattern : {--,--,--,--,--}

ex:

variation 1 = {01,02,03,04,05}

variation 2 = {01,02,03,05,04} // Invalid variation. Already exist with different order.

variation 3 = {01,02,03,04,06}

variation 4 = {34,56,73,34,98} // Invalid variation. Contains repetition(34).

.

I found this answer for the non-repeatable variations but how to apply non-order considered variations to this formula?

Thank you.

https://math.stackexchange.com/a/35180/166912

kub0x · Accepted Answer · 2017-02-12T09:48:24.327

1

Since you are not considering order and repetition then what you are looking for is the typical Combination, this is, how to form groups of numbers from $0$ to $99$ taking those numbers in $5$.

Formally, $C^{100}_5 = \frac{100!}{5!(100-5)!} = 75287520$ possible arrangements.

If you take order in consideration then instead of a Combination you have a Variation where each aforementioned group is permuted, so you will end up with a total of:

$\frac{100!}{(100-5)!} = 9034502400$ possible arrangements.

edited Feb 12 '17 at 09:48

answered Feb 12 '17 at 08:49

kub0x

2,117
1
15
25

What if different orders are acceptable? – Eftekhari Feb 12 '17 at 08:55
@Eftekhari: I have edited my answer to include ordering. – kub0x Feb 12 '17 at 09:07
I think you have a typing mistake in the second part. It should be (100 - 5)!. – Eftekhari Feb 12 '17 at 09:46
@Eftekhari: Yes you are right. It is always "n - k" where "n=100" and "k=5" so it is "(100-5)!". Fixed it. – kub0x Feb 12 '17 at 09:48

Variation number of tw0 digit numbers in array of 5 without repetition and order

1 Answers1