Find the sum of all 4-digit numbers formed by using the digits 1, 2, 3, 4, 5

Question

...and where no digit is to be repeated in any 4-digit number.

Yes, I am aware of similar questions on this site, but none of the answers gave me insight as to how this works.

I would like an answer in ridiculously simple terms, like explaining it to a 5 year old.

One answer:

Total numbers formed using those 5 digits is 5P4. i.e. 120 numbers.

Since all digits will equally occupy the unit's place, therefore 1 will occur 24 times in unit's place. Similarly 2 will occur 24 times in unit's place. So on and so forth. Therefore sum of all digits in unit's place will be equal to 24 x (1+2+3+4+5)=24 x 15.

Similarly sum of all digits in ten's place will be equal to 24 x 150.

Therefore total sum of all numbers =24 x (15+150+1500+15000)=399960 .

Why is there the sum (1 + 2 + 3 + 4 + 5)? Am I misreading the question? Does "no digit is to be repeated in any 4-digit number" mean that there shouldn't be a number like 4432, or does it mean that a number should not be repeated in the same "unit slot"?

Answer 1

Your answer is trying to say that when you write out all 120 numbers, it will look something like this:

1234

1235

...

2134

2135

...

5123

5124

There will be 24 ones in the thousands place, 24 twos in the thousands place, etc. This is because once we set a certain number in the thousands place (e.g. one), then we have 24 distinct four digit numbers.

So we calculate the sum of the 120 valid 4 digit numbers by first looking at the sum of the digits in the thousands place. This sum is $$1000 * 24 * 1 + 1000 * 24 * 2 + 1000 * 24 * 3 + 1000 * 24 * 4 + 1000 * 24 * 5$$

The 1000 comes from looking at the sum of the digits in the thousands' place. The solution you provided factored this result as $1000 * 24 * (1+2+3+4+5)$.

Now that we've calculated the sum of the digits in the thousands' place, we can do the same for the hundreds' place, tens' place, and ones' place similarly:

Hundreds' Place: $100 * 24 * (1+2+3+4+5)$

Tens' Place: $10 * 24 * (1+2+3+4+5)$

Ones' Place: $1 * 24 * (1+2+3+4+5)$

Adding these values together, we get the answer, which is 399960.

Answer 2

It's just a shorthand for adding up the $120$ possible values of the ones place.

The digit $1$ occurs $24$ times, as does $2, 3, 4,$ and $5$,

If your digits were $(2,4,5,7,9)$ then the sum would be $24 \times (2+4+5+7+9)$.