Permutation without repetition/ Solution result doesn't match! this is a study question given to us by the professor, but i am a bit confused regarding the result i get, it doesn't match with the professor's one! Is there anything that i am doing wrong and missing here, if so please correct me.
Thank You. With Respect Umer Selmani
the question:
How many numbers larger than 56100 can formed with digits from (0,1,5,6,7), repetitions not allowed? 
the solution given to us: $2 * 4! + 1 * 1 * 3! + 1 * 1 * 2 * 2!$
the solution i got:       $2* 4! + 1 * 1 * 3! + 1 * 1 * 1 + 2!$
 A: We have


*

*for $7xxxx$ or $6xxxx\implies2\cdot 4!$

*for $57xxx\implies1\cdot 1\cdot 3!$

*for $561xx$ or $567xx\implies1\cdot 1\cdot2\cdot 2!$
A: I will give a very explicit solution, but note it is too cumbersome for such a simple problem which is not best practice..
You want numbers larger than (strictly larger than) $56100$, and the digits you have available to you are $0$, $1$, $5$, $6$ and $7$ with repetition of digits not allowed. Let us denote 
$$ 56100 = d_1 d_2 d_3 d_4 d_5 $$
So the first observation, since repetition is not allowed, is that all of your numbers will have be five digits long
We can enumerate all possibilities by cycling through digit by digit:


*

*Numbers whose first digit is larger than $d_1$: For this case, the first digit can either be 6 or 7, two choices.  Then for each choice, you can permute the remaining digits, $4!$ ways.  All together, $2\cdot 4!$ ways

*Numbers whose first digit is $d_1$ and second digit is larger than $d_2$: Well there is only one choice for the first digit, $5$.  Which leaves one choice for the second digit, $7$.  And the rest of the digits you can permute any way you like, in total $1\cdot 1\cdot 3!$ ways

*Numbers whose first two digits are $d_1 d_2$ and last three digits larger than $d_3 d_4 d_5$: Well the first two digits are fixed, so each has one way to do that.  And since $d_3 d_4 d_5 = 100$, we only need the third digit to match or exceed $d_3$.  there are two choices, and the remaining digits are permuted any way.  In total $1\cdot 1\cdot 2\cdot 2!$ ways


Combining all 3 cases, result is
$$ (2\cdot 4!) + (1\cdot 1\cdot 3!) + (1\cdot 1\cdot 2\cdot 2!)$$
A: The numbers from $0,1,5,6,7$ that are greater than $56100$ all have 5 digits.
There are $120$ arrangements of them, we can ignore the $24$ that start with a 0 and the $24$ that start with a 1. Also the $6$ that start with 50 and the $6$ that start with 51, and the $2$ that start with 560.
So $120-48-12-2=58$.
