They are asking for number of solutions where x1>= 1 and x2,x3,x4 <=6
Method 1: Inclusion-Exclusion
Lets find total solutions where x1>= 1.
So, x1 + 1 + x2 + x3 + x4 = 15
So, that comes as $\binom{17}{3}$ = 680 ............(1)
Now, we need to find solution where (x2>=7 or x3>=7 or x4>=7).
Lets denote such events as E(x2>=7 or x3>=7 or x4>=7).
$\therefore$ E(x2>=7 or x3>=7 or x4>=7)
= E(x2>=7) + E(x3>=7) + E(x4>=7) - E(x2>=7 $\cap$ x3>=7) - E(x4>=7 $\cap$ x3>=7) - E(x4>=7 $\cap$ x2>=7) ..........(2)
(only 2 of x2, x3, x4 can be more than 7 at the same time)
Now, when x2>=7, then equation will become x1 + 1 + x2 + 7 + x3 + x4 = 15
Thus, E(x2>=7) = $\binom{10}{3}$ = 120
$\therefore$ E(x2>=7 or x3>=7 or x4>=7) = 120 + 120 + 120 - 1 - 1 - 1 = 357
Thus, answer = 680 - 357 = 323
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Method 2: Generating functions
For the given equation, generating function is as follows:
(x1 + x2 + .......+ x15) ( 1+ x1 + .....+ x6) ( 1+ x1 + .....+ x6) ( 1+ x1 + .....+ x6)
We have to find coefficient of x15 in the above equation.
So, above equation becomes
= (x1 + x2 + .......+ x15) ( 1+ x1 + .....+ x6)3
= x1(x0 + x1 + .....+ x14)( 1+ x1 + .....+ x6)3
= x1 * $\frac{1-x^{15}}{1-x}$ * $(\frac{1-x^{7}}{1-x})^{3}$
Now, we cancel $x^{15}$ in the second term because we need max of x14 from second and third terms.
So, above equation becomes
= x1 * $(1-x^{7})^{3}$ * $(1-x)^{4}$
= x1 * (1 - 3x + 3x14) * ( 680x14 + 120x7 + 1)
I have selected only those components out of expansion for $(1-x)^{4}$ which will give me x14 out of second and third components above.
So, just calculating the coefficients of x14 from above equation,
answer = 323