How to count possibilities of 10 pics in 10 albums?

Question

Sorry if I won't sound too easy. I'd like to know how could I find out the exact number of the total possibilities of the next problem.

There are 10 different pictures which are appearing in 10 albums. The same picture cannot be in the same album multiple times. But the same picture can be used in all of the albums (at the same time). How many different albums are possible if we count as a different one when the position is changed of any pic. All of the albums have different size of free places. (All of the places must be filled every time.)

In the 1st album: only 1 picture can be (is) placed (it can be any of the 10 pictures - so this album can have 10 possibilities, here there is no meaning of the order of the single pic, so this one is easy. We are at 10 after this stage.)

In the 2nd album: 2 pictures are in this (any 2 from the 10, also they can be in 2 different orders: A,B or B,A on each pairs)

In the 3rd album: 3 pictures must be placed here (take 3 pics from the 10, take the number of the different orders possible and somehow combine these, then repeat for all possible 3-pic-selections)

etc... up to the last, 10th album where all of the pics are present (here there is only 1 possibility of the selected pics (all of em comes 1-10), but because of the different possible orders we must get some big number here too)

I believe, first need to count all of the 10 albums separately, then just adding their values at the end.

What do I need to know to be able to solve this? Factors too, I guess but cannot figure out how to do it properly. Thank you for your time!

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EDIT: Adding some sample calculations to my post as I think I was not clear enough.

Only 1 album can exist at a time. It can be any size (from 1, up to the number of available pics number). How many possible ways can appear this album?

If I'd have 4 pictures and albums I'd have this total possibilities (what I am searching):

1
2
3
4

1 2
1 3
1 4
2 1
2 3
2 4
3 1
3 2
3 4
4 1
4 2
4 3

1 2 3
1 2 4
1 3 2
1 3 4
1 4 2
1 4 3
2 1 3
2 1 4
2 3 1
2 3 4
2 4 1
2 4 3
3 1 2
3 1 4
3 2 1
3 2 4
3 4 1
3 4 2
4 1 2
4 1 3
4 2 1
4 2 3
4 3 1
4 3 2

1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1
_______

4 + 12 + 24 + 24 = 64

If I didn't miss anything it seems the amount of the 2nd is 3x of its previous, the 3rd is 2x the prev, and the 4th is the same as its prev. Strange! So for 4 (instead of 10) the answer should be 64.

Answer 1

You can make an album with $p$ pictures from the $10$ in $$ A(p) = 10 \times 9 \times 8 \times \cdots \times (10-p+1) $$ ways, since order matters and that's how you pick the pictures in order. (That's $10$ ways for album $1$, $90$ ways for album $2$, $720$ ways for album $3$, and so on.)

Since the albums have nothing to do with one another, to find the total number of different albums you multiply to get $$ A(1) \times A(2) \times \cdots \times A(10) . $$

(Multiply instead of add because for each possible first album you can have any of the second album and so on. It's why there are $12$ possibilities if you roll a die an flip a coin.)

To make the calculation, note that $$ A(p) = \frac{10!}{p!} =\frac{10 \times 9 \times \cdots \times 1}{p \times (p-1) \times \cdots \times 1} $$ and the google calculator knows about factorials: I entered

(10)*(10*9)*(10*9*8)*(10!/7!) 

and got

466560000

Then

(10)*(10*9)*(10*9*8)*(10!/7!)*(10!/6!)*(10!/5!) 

was

7.1108223e+16

but for more factors Google gave up. If you really need an exact answer someone may write a program (Python would be easy) and post the result here.

Edit:

I rearranged the computation to collect up the powers of each of the integers between $2$ aand $10$. Then Wolfram alpha had no problem with

(10^10)*(9^9)*(8^8)*(7^7)*(6^6)*(5^5)*(4^4)*(3^3)*(2^2) 

returning

215779412229418562091680268288000000000000000

or

$$ 2.15779412229418562091680268288 \times 10^{44} $$