How many combinations of 2 numbers are there in 5 numbers?

Answer 1

It depends on whether the number order can or cannot be identical. Also whether you can use the same number twice. There are 10 unique pairs possible (i.e. 1 and 5 is the same as 5 and 1), but 20 sequential pairs where the order of the pair makes them different, and 25 if you can use the same number in a pair.

10 = 1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, 4-5

20 = (include the reverse of those)

25 = (include 1-1, 2-2, 3-3, 4-4, 5-5)

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A combination is a selection of a number of items from a group without regard to order (i.e. AB = BA). Since order is not important, then the number of combinations is given by the formula: n!/(n - r)! r!

where n = the total number of items. r = the number of items chosen from n

(recall that n!, pronounced n factorial, means the continued product of successive natural numbers from 1 to n. So 5!= 5x4x3x2x1, for example)

Let the 5 numbers be 1, 2, 3, 4, and 5. Possible combinations are: 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, 3 and 4, 3 and 5, 4 and 5.

(Note how '2 and 1' is not listed because '2 and 1' is the same as '1 and 2' when using combinations).

So there are 10 combinations of 2 numbers taken from 5 numbers.

Using the formula there are 5!/3!2!=10 ways to pick 2 items from 5 where the order does not matter. We also assume we are not replacing an item after we pick it.

When we pick r items from n with replacement, we must use a different method or another formula. We often use a method known as bars and stars. Alternatively, we can use the formula (n+r-1)! / r! (n - 1)!. When we put the items back, we can pick more than n items. So it would make sense to talk about how many ways to pick 7 items from 5, whereas without replacement this is impossible.