Generating Function

Question

Q.find number of 7 digit number with sum of digits  equal to  11 and formed using digits 1 ,2 ,3 Can we do this with help of generating functions.

Comments

is answer 161??

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Recurrence : (1-x³)⁷ (1-x)^-7

To find : [x⁴]
which is 10c4 - 7*7c1
=161

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@saxena0612  [X⁴] or [x¹¹] ?

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after building recurrence you wil get x⁷ therefore we just need x4 in order to find x¹¹

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=( x+x²+x³ )⁷

=x⁷(1+x+x²)⁷  correct ?

why didn't form the recurrence as (1+x+x² +x³)⁷ ?

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@junaid Ahmad 

<1,2,3 > are given sequence??

if yes then

1+2x+3x$^{2}$

otherwise anyone explain

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Here 3 elements are there 1,2,3
So, we have taken $\left ( x+x^{2}+x^{3} \right )$
And as they are all 7 digit number , So, power is taken 7 i.e. $\left ( x+x^{2}+x^{3} \right )^{7}$

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@srestha

Can I choose (1+x+x$^{2}$)???

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No
it cannot be
if 1st element power is 0, that means atleast 1 element in every sequence should not exists
See the selected ans
then (3,2,2,1,1,1,1) where all 3 elements present doesnot exists

Answer 1

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!}$ =$161$ 
 

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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}$
[$x^4$] $10_c4$ $- 7\times 7_c1$ 
$=161$

Comments

@saxena0612

How to write for any place function is given as x+x$^{2}$+x$^{3}$ ???

why you not write

<1,2,3 > are given sequence??

if yes then

1+2x+3x$^{2}$

otherwise, anyone explains