search
Log In
1 vote
162 views
How many positive integers less than 1,000,000 have the sum of their digits equal to 19?
in Combinatory
closed by
162 views
1

I am little doubtful with regard to below points.

The equation I understood is

d1 +d2 + d3+ d4+d5+d6=19

where each  0$\leq$di$\leq$9 (Actually answer comes when  I take each di in this range)

my query was why not each di will be in range 1$\leq$di$\leq$9

because question mentions for positive integers so each digit must be from 1 to 9.

0
yes,that will be equation

ans 42504?
0
@Ayush it's because 1 digit number can be written as 00000 or 000001 or 0000002 or so on..

I mean, 000 289 = 19 is also a possibility.
0
@Srestha-No it's 30492.

I think Why di ranges from 0 to 9 because

Firstly the question says

"the sum of digits must be 19"

so with all zeroes, I cannot have sum as 19 (so obviously there would be a positive integer whose sum of digit would be 19)

and since with up to 6 digits, each digit can range from 0 to 9 only that's why each di is from 0 to 9.

Related questions

26 votes
3 answers
1
4.9k views
Find the number of integral solutions of $\large 2x + y + z = 20$ with $x, y, z >= 0$ ? I tried finding coefficient of $x^{20}$ in $(1 + x^2 + x^3 + ...... + x^{10})(1+ x^2 + .......... + x^{20})^2$, but it gives wrong answer ?
asked Sep 1, 2016 in Combinatory mcjoshi 4.9k views
11 votes
1 answer
2
2 votes
1 answer
3
508 views
Use Generating function to determine,the number of different ways $10$ identical balloons can be given to four children if each child receives atleast $2$ ballons? Ans given $(x^{2}+x^{3}+.........................)^{4}$ But as there is a upper bound I think answer will be $(x^{2}+x^{3}+.........................+{\color{Red} {x^{10}}})^{4}$ Which one is correct? plz confirm
asked Dec 4, 2018 in Combinatory srestha 508 views
1 vote
1 answer
4
...