Number Game

We have that x+2y+3z=10 and x+y+z=7, where x y z are the number of "1", "2" and "3" respectively. Then x=4+z and y=3-2z where the only meaningful solutions are (4,3,0) and (5,1,1). For z=0 (ex. 1111222) we have 7!/(4!3!) diferent ways of sorting the digits and for z=1 (ex. 1111123) we have 7!/5! different combinations. Then we have a total of 77 different numbers that fulfill the given conditions

This problem is identical to the problem of dividing 10 identical balls into 7 urns where each urn contains at least 1 ball, and at most 3. We can preplace 1 ball in each urn, and are left with dividing 3 balls over 7 urns, where we can place at most 2 balls in one urn. With stars and bars we have ${3+7-1 \choose 3} = \boxed{84}$ of such distributions, where we need to subtract distributions in which we put 3 in one urn. There are $\boxed{7}$ of these. In total we have $84 - 7 = \boxed{77}$ distrubutions.

It is a problem of combinations!

There are only two ways of forming an seven digit positive integer whose sum is equal to 10 and using digits 1,2 and 3

They are : (i)five 1's, one 2, one 3 (ii)four 1's, three 2's

Now the combinations for (i) will be "(7C5)x2" and for (ii) will be "(7C4)x1" Adding you will get 35+42=77!

just coded.. and have got the answer... :P

Amongst the solutions of the Diophantine equation a+2b+3c=10 we seek those where a+b+c=7, then the sum of 7!/a!b!c! for each solution.

In Python:

>>> from sympy.solvers.diophantine import diophantine

>>> from sympy import *

>>> a,b,c=symbols('a b c ')

>>> diophantine(a+2 b+3 c-10)

set([(-3 c + 2 t + 10, -t, c)])

>>> for C in [0,1]:

... for B in range(-5,2):

... A=-3 C+2 B+10

... if A-B+C == 7 :

... A,-B,C, A+2 (-B)+3 C, factorial(7)/(factorial(A) factorial(-B) factorial(C))

...

(4, 3, 0, 10, 35)

(5, 1, 1, 10, 42)

>>> 35+42

77