Question

How many pairs can be made?

Suppose we have a vector starting with 1 2 3.
and we also have 2 spare values, 4 and 5 that are not initially in the vector. the numbers can be in any order and no duplicates. How many pairs of 3 digits can we have?


Pairs I have came up so far.

123 (45) <-- spare ones
423
143
124
523
153
125
453
145
425

Answer 1

You need to choose \(3\) different digits from a set of \(5\) digits.
Total number of ways of doing this is given by :
\[\binom{5}{3}\]

Answer 2

we have 5 options for the first choice

once we pick one, we have 4 options for the next choice: 5 choices with 4 options each is 5*4 = 20 picks so far

once we pick the second one, we have 3 options remaining: 20 choices with 3 options each is therefore 20*3 = 60 picks in total

but theres only one flaw with my strategy here, one that i foresaw and can adjust for.

abc = acb = bac = cab = bca = cba

each options has 6 arrangements that are all being counted as different, instead of the same. my count is 6 times to big, so lets adjust it by dividing by 6

60/6 = 10