How many four digit numbers can be formed by the use of digits 1 2 3 4 and 6 without repeating any digits?

Answer 1

There are 120.

The total of possible combinations of n numbers is n! (n factorial)

equal to (n)(n-1)(n-2)...(1)

For 4 numbers, this is 24 combinations, and for 5 numbers 120 combinations.

Taking only x digits, the number of combinations is reduced to only the first x values of the factorial.

There are 5 one-number combinations

There are 5 x 4 two-number combinations

There are 5 x 4 x 3 three-number combinations

There are 5 x 4 x 3 x 2 four number combinations

So that for a combination one digit less than n, i.e. x = n-1, the number is still n!

If you write out the 24 combinations of 1, 2, 3, and 4, you can see that the number 6 (or any number) could be placed into the numbers in either the first, second, third, or fourth digit, creating 4 sets of 24 new numbers, for a total of 120.

They are:

1234 2134 3124 4123 6123

1236 2136 3126 4126 6124

1243 2143 3142 4132 6132

1246 2146 3146 4136 6134

1263 2163 3162 4162 6142

1264 2164 3164 4163 6143

1324 2314 3214 4213 6213

1326 2316 3216 4216 6214

1342 2341 3241 4231 6231

1346 2346 3246 4236 6234

1362 2361 3261 4261 6241

1364 2364 3264 4263 6243

1423 2413 3412 4312 6312

1426 2416 3416 4316 6314

1432 2431 3421 4321 6321

1436 2436 3426 4326 6324

1462 2461 3461 4361 6341

1463 2463 3462 4362 6342

1623 2613 3612 4612 6412

1624 2614 3614 4613 6413

1632 2631 3621 4621 6421

1634 2634 3624 4623 6423

1642 2641 3641 4631 6431

1643 2643 3642 4632 6432