GATE CSE
First time here? Checkout the FAQ!
x
+4 votes
391 views
How many positive integers less than 1,000,000 have the sum of their digits equal to 19? (using generating function)
asked in Combinatory by Veteran (12.7k points)   | 391 views
$\binom{24}{19} - 6*\binom{14}{9} = 30492$
pls give detail approach . yes ans is correct.

1 Answer

+13 votes
Best answer
  • The number is $<$ $1000000$ , $\Rightarrow$ it contains 6 digits.
  • Each of these digits can be one of $0,1,2,3....9$

$\Rightarrow$ problem reduces to no of integral solution to the following equation

$d_1+d_2+d_3+d_4+d_5+d_6 = 19$ where $0\leq d_i \leq 9$

Using generating function : (how to use)

$$\begin{align*} & \ \ \ \left [ x^{19} \right ]\left ( 1+x+x^2+x^3+....+x^9 \right )^{6} \\ &=\left [ x^{19} \right ]\left [ \frac{1-x^{10}}{1-x} \right ]^{6}\\ &=\left [ x^{19} \right ]\left ( 1-x^{10} \right )^{6}. \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \\ &=\left [ x^{19} \right ]\left [ \sum_{r=0}^{6}.\binom{6}{r}.\left ( -x^{10} \right )^{r} \right ]. \left [ \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \right ] \\ &=\left ( -1 \right )^{0}*\binom{24}{19} + \left ( -1 \right )^{1}*6*\binom{14}{9} \\ &=30492\\ \end{align*}$$

NOTE:

1. $1+x+x^2+x^3+.....x^n = \frac{1-x^{n+1}}{1-x}$

2. $\frac{1}{(1-x)^n} = \sum_{r=0}^{\infty}\binom{n+r-1}{r}.x^r$

3. $\left [ x^{19} \right ]$ means coefficient of $x^{19}$ of the whole expression.

answered by Veteran (45.6k points)  
edited by
thanks..:-)

and can you share some good link for studying generating functions because i have almost no idea about them and all these questions of finding coefficents are using these functions.
Like all would agree on Kenneth for gate. I also think kenneth generating function + generalized permutation and combination chapter. There is also a very good pdf ( not long) only on gen function. Go though the how to use link in this post and in the new page ..there  in the first paragraph page15 link.
Why did you put 1 in first and 19 in latter and not the other way round?

Debashish Deka  (Convolution Rule). Let A(x) be the generating function for selecting items from set A, and let B(x) be the generating function for selecting items from set B. If A and B are disjoint, then the generating function for selecting items from the union A∪B is the product A(x) B(x).

Since value of each di is in set {0,1,2,3,4,5,6,7,8,9} .How are d1,d2,...,d6  disjoint ?? Plz explain.

 



Top Users Apr 2017
  1. akash.dinkar12

    3508 Points

  2. Divya Bharti

    2542 Points

  3. Deepthi_ts

    2040 Points

  4. rude

    1966 Points

  5. Tesla!

    1768 Points

  6. Shubham Sharma 2

    1610 Points

  7. Debashish Deka

    1588 Points

  8. Arunav Khare

    1454 Points

  9. Kapil

    1424 Points

  10. Arjun

    1420 Points

Monthly Topper: Rs. 500 gift card

22,076 questions
28,040 answers
63,230 comments
24,135 users