Number for ways to arrange 10 distinct object into 4 distinct boxes

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What is the number of ways to arrange 10 distinct objects into 4 distinct boxes, where each box hold no more than 4 objects

This question was asked earlier today, but has been deleted, I don't know why

I attempted to solve the problem, and came onto this

Put $10$ distinct object in $4$ distinct boxes, where a box cannot hold more than $4$ object

Distinct object say- $(1,2,3,4,5,6,7,8,9,10)$

Distinct boxes say- $(x_1,x_2,x_3,x_4)$

So generally there are $16$ spaces in all box, to which we want to fix in $10$ distinct objects so some spaces might be left

There are two ways of doing this

IF THE OBJECTS ARE NOT REPEATED

If the object are not repeated, there would be $6$ spaces each time

I'll call this the Occurance of $6$ space ( no extra object )

So the number of ways of doing this is $10!×16C10 = 10!×8008$ ways

IF THE OBJECTS CAN BE REPEATED

If the object can be repeated, the $6$ spaces would give rise to other objects ( repeatition ) and the remaining space, More importantly let's not forget to include all the possible combination w.r.t repeating of the objects

Occurance of $1$ object and $5$ space

Occurance of $2$ object and $4$ space

Occurance of $3$ object and $3$ space

Occurance of $4$ object and $2$ space

Occurance of $5$ object and $1$ space

Occurance of $6$ objects ( no extra space )

So that the total number of ways to scatter the district object into the distinct boxes in repeatition, is the sum of the number of ways here

Therefore $10!×16C10×6×10 + 10!×16C10×6×(10+10C2) + 10!×16C10×6×(10+10C2+10C3) + 10!×16C10×6×(10+10C2+10C3+10C4) + 10!×16C10×6×(10+10C2+10C3+10C4+10C5) + 10!×16C10×6×(10+10C2+10C3+10C4+10C5+10C6)$

Which becomes

$10!×16C10×6×(10*6+10C2*5+10C3*4+10C4*3+10C5*2+10C6*1)$

$10!×16C10×6×2109$

The total number of ways in scattering the distinct objects is the sum of it ways in repeatition plus it ways in non-repeatition, the solution then becomes

$10!×16C10 + 10!×16C10×6×2109$

$10!×16C10×(1+12654) = 10!×16C10×12655$

$9,643,717,440,000$ ways

Have I done it right, or what's wrong with this approach, thanks for helping