Given : Seven digit number , no of digit allowed :1,2,3,Sum should be 11
To find : No of such combination :
Solution :
1) Without generating function :
$ (3,3,1,1,1,1,1) , (3,2,2,1,1,1,1) , (2,2,2,2,1,1,1)$
No of combinations : $ \frac{7!}{2!5!} + \frac{7!}{2!4!} + \frac{7!}{3!4!}$ =$161$
2) Generating function:
for any place function given as :$x+x^2+x^3$
we have 7 of this :
function $=(x+x^2+x^3)^7$
[x11] $x^7$ $(1+x+x^2)^7$
[$x^4$]$(1-x^3)^7$$(1-x)^{-7}$
[$x^4$] $10_c4$ $- 7\times 7_c1$
$=161$