Set Interpretation of number of ways 6 different objects put into 10 different cells Question from book: In how many ways can 6 different objects be put into 10 different cells such that in a cell there won't be more than one object.
Answers:
Method 1: In every cell there will either be one object or there won't be any object. We'll choose the 6 cells into which we will put the objects. This can be made in $ { 10 \choose 6 } $ ways. Since the objects are different, then the ordering between them matters. Thus the number of possibilities to put the 6 objects in 10 cells is:  $ { 10 \choose 6 } \cdot 6! = \frac{10!}{4!} $
Method 2: Order all the objects one after the other into the different cells: for cell #1 - 10 possibilities , for cell #2 - 9 possibilities and so forth. So by multiplication principle, the number of possibilities is $ P(10,6) = \frac{10!}{4!} $
My difficulty:
I solved the problem using method 1 but when I tried to write the sets participating in the multiplication and the resulting set then things looked weird, here's what I did:
$ { 10 \choose 6 } $ represents the set $ \alpha =  \{ \{1,2,3,4,5,6 \} ,\{1,2,3,4,5,7 \},\{1,2,3,4,5,8 \},... \} $ where the numbers from 1 to 10 are the indices of the 10 different cells.
$ 6! $ represents the set $ \beta =  \{ (ABCDEF) , (ABCDFE) , (ABCFDE) ,... \} $ where the symbols A,B,C,D,E,F represent the 6 different objects.
By multiplication rule $ \gamma =  \alpha \times \beta = \{ ( \{1,2,3,4,5,6 \} , (ABCDEF) ) , ( \{1,2,3,4,5,6 \} , (ABCDFE) ) ,... \} $ ~ $ { 10 \choose 6 } \cdot 6! $
However, looking at the elements of the set $ \gamma $, they are ordered pairs and I can't make sense out of them, they don't look like they're representing to me lists of 6 objects ( lists because the objects are different so order matters ) with the fact that they're put in different cells. However I was thinking maybe the elements represent to me some sort of bijections from the first element in the ordered pair to the second? ( for example: maybe the first ordered pair in $\gamma $ represents a bijection from $ \{1,2,3,4,5,6 \} $ to $ (ABCDEF) $ ? ).
My questions:
So essentially, I'm asking; what does the set $ \gamma $ above represent to me?, am I right that the elements of $ \gamma $ represent bijections from 1st element ( which is a set ) in the ordered pair to 2st element? What the set of the "number of ways" of 6 different objects put into 10 different cells should look like?
 A: The element $\{1,2,3,4,5,6\},(ABCDEF)\}$ is just the encoding of the fact that object $A$ goes to box 1, $B$ to $2$ etc.
You're just counting the number of injections (1-1 maps) from the set of objects to the set of boxes, in essence. So the number of $6$-combinations out of $10$, or indeed $\binom{10}{6}\times 6!$. Or,  the first object has $10$ options, the second $9$ etc so we get $$10\times 9 \times 8\times 7 \times 6 \times 5$$ options directly, but in a calculator that only has a binomial function, it's handy to write it as that product for computation purposes. We don't need to see that specific product as the result of some product rule, though I showed that we could, if we wanted to.
A: The $abc$ component can be used to transform the $123$ component into a unique result.
This is the complete table for placing $3$ distinct objects into $4$ distinct cells, object $k$ is placed into cell $c_k$.
$214$ for example meaning:

Place object $1$ into cell $2$, object $2$ into cell $1$ and object $3$ into cell $4$.






abc
acb
bac
bca
cab
cba




123
123
132
213
231
312
321


124
124
142
214
241
412
421


134
134
143
314
341
413
431


234
234
243
324
342
423
432



