How to count the number of combinations of a string consisting of 10 characters, where only Latin letters in lower case and numbers are used.

I’d love to know the formula.

## Answer 1, authority 100%

Total characters – 36.

If repetitions can be – in the first place, how many ways can you choose a character? 36. For each first, how many can you choose the second? 36. Total – 36 * 36. For every first two … – well, and so on.

Total – 36 ^{10 }.

For an alphabet of N characters and line length m –

If there can be no repetitions, then only 35 are in second place (one has already been selected), in third – 34 (two have already been selected) … And so on. Total – 36 * 35 * 34 * 33 * 32 * 31 * 30 * 29 * 28 * 27 = 36! / 26 !.

For an alphabet of N characters and line length m – the number of *placements *

## Answer 2, authority 27%

Two formulas can be used.

If characters can be repeated, then any of 10 characters can take one 36 values (26 Latin letters plus 10 digits). We can say that this is a 10-digit number in the 36-ary number system. The number of combinations will be 36 ^{10 }or 3.6561584 × 10 ^{15 }.

If symbols cannot be repeated, then we are dealing with *placements *. There are also *combinations *, but in this case they are not suitable because the placement of `123abc `

and `abc123 `

will be different, but the combination will be the same same.

The number of placements from *n *to *k *is calculated using the formula A ^{k }_{n }= n! / (n – k) !, that is, in your case it will be 36! / 26! or 9.2239326 × 10 ^{14 }

## Answer 3, authority 9%

The alphabet used contains 36 characters: 26 letters and 10 numbers.

Number of placements without repeats = n! / (n-k)! = 36! / 26!

Number of placements with duplicates = n ^{k }= 36 ^{10 }

## Answer 4

number of combinations = C!

Ie in your case 10! combinations …