SOLUTION: In how many ways can the digits in the number 7,633,333 be​ arranged?

If it is assumed that the value of this number remains unchanged, then there is only one arrangement.

It is what you see.


If any permutations are allowed and we do not care about the value of the number, 

then we have 7 positions, in all, and 3 (three) different digits.

The digit 3 is repeated five times.


So, the number of different distinguished arrangements is  

     = 6*7 = 42 after reducing the fraction.    ANSWER



It can be easily explained. We can place the digit 7 in any of 7 positions
and we can place the digit 6 in any of 6 remaining positions.
After that we place the "3's" to fill the 5 remaining positions - it produces one arrangement.
Doing and counting this way, we get 7*6 = 42 different distinguishable arrangements.

There are 7 positions to place the 7.
For each of those 7 positions to place the 7, there are 6 positions
to place to 6.  

That's 7x6 = 42. (We put 3's in the other 5 places)

Or we can use the formula:

         

The number of things (digits) to be arranged is 7 because 7633333 has 7
digits.  Only the 5 3's are indistinguishable.  So the answer this way is



Here are all 42.  7 rows of 6 each:

7633333, 7363333, 7336333, 7333633, 7333363, 7333336,
6733333, 6373333, 6337333, 6333733, 6333373, 6333337, 
3763333, 3736333, 3733633, 3733363, 3733336, 3673333, 
3637333, 3633733, 3633373, 3633337, 3376333, 3373633, 
3373363, 3373336, 3367333, 3363733, 3363373, 3363337, 
3337633, 3337363, 3337336, 3336733, 3336373, 3336337, 
3333763, 3333736, 3333673, 3333637, 3333376, 3333367. 


Edwin