How many 3-digit numbers can be formed using the digit 123456 and 7 if no digit can be repeated in number?

Answer 1

With questions like these it is often difficult to remember whether we are looking for permutations or combinations. In English we use these two terms interchangeably but in mathematics we must be more precise. The way to remember is that if the order matters, then position matters, thus position = permutation. If order does not matter then we're looking for combinations. Put simply, 123 and 321 are the same combination but different permutations. In this case we're looking for permutations (thus combination locks should really be called permutation locks).

The formula for calculating permutations is fairly simple: given 7 digits from which we must select 3, we have 7^3 = 7 x 7 x 7 = 343 permutations in total. However, in order to exclude repeated digits we must use a different formula. When we select the first digit we have 7 choices (just as we do with repetition allowed), but when we choose the next digit we only have 6 choices remaining and for the final digit we only have 5 choices remaining.

To express this more mathematically, we need to use factorials. Factorial 7 (written 7!) is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. This tells us there are 5040 different ways we can write all the digits from 1 to 7 without repetition. In order to calculate all the 3-digit permutations we need to stop multiplying when we reach 4. To achieve this, we use the following formula:

n! / (n-m)!

where n is the number of elements in the set and m is the number of elements we wish to select.

Thus we get:

7! / (7-3)! = 7! / 4!

If we (partially) expand this we find we have:

7 x 6 x 5 x 4! / 4!

The 4!s above and below the division will cancel each other out, so we get:

7 x 6 x 5 = 210 permutations without repetition.

If we were to fully expand this out we find we have:

(7 x 6 x 5 x 4 x 3 x 2 x 1) / (4 x 3 x 2 x 1) = 5040 / 24 = 210 permutations without repetition.

Note that if we were only interested in the number of combinations (without repetition), we divide the number of permutations by 3! For that we use the following formula instead:

n! / (m! x (n-m)!)

Thus we'd get:

7! / (3! x 4!) = (7 x 6 x 5 x 4!) / (3 x 2 x 1 x 4!)

The 4!s cancel each other out so we get:

(7 x 6 x 5) / (3 x 2 x 1) = 210 / 6 = 35 combinations without repetition.

Answer 2

There are 35 unique combinations of subsets of 3 from a set of 7:

123 124 125 126 127 134 135 136 137 145 146 147 156 157 167234 235 236 237 245 246 247 256 257 267

345 346 347 356 357 367

456 457 467

567

Each combination also has 6 permutations:

E.g., the first combination has the following permutations: 123, 132, 213, 231, 312, 321.

Therefore, 6 x 35 = 210 possible 3-digit numbers where no digit is repeated.