# DM-Sets,Relations and Functions-Q8

How many nonzero entries does the matrix representing the relation R on consisting of the first 1000 positive integers have if R is asked Jul 5, 2019
reshown Jul 6, 2019

matrix representation of  will contain  cells.

can also write above equation with dummy variable c,

answered Jul 6, 2019 by (1,730 points)
selected Jul 9, 2019
i have done  like this .
i also did the same...
me too
same ans
Can u tell me where i went wrong i approached this question in this way ,,

We have condition a+b<=1001 so possible equations are as follows
a+b=2    total cases here 2-1+2C2 =3C2 =3
a+b=3    total cases here 2-1+3C3 =4C3 =4
.....
.....
a+b=1001 total cases here 2-1+1001C1001 = 1002C1001=1002

if we sum all these we get 502500
bro i dont know how you are getting a+b =2 in 3 ways. a+b=2 only when a=1,b=1;
similarly a+b=2 only when(1,2) or (2,1) so only two ways. similarly for the rest.
your solution for a+b=x is wrong
thanks i think i missed the condition a>=1 and b>=1
hmm, you missed something. (a, b) can only be the given set A={1,2,...,1000}
Thanks bro it should be like this
it can be solved by using the concept of combinatories,
We have condition a+b<=1001 so possible equations are as follows
a+b=2    since a>=1 and b>=1 modfied eq is a+b=0 total cases here 2-1+0C0 =1C0 =1
a+b=3    similarly here it is a+b=1  and total cases are  2-1+1C1 =2C1 =2
.....
.....
a+b=1001  similarly here it is a+b=999  and total cases are  1000C999 =1000

total cases here 1 + 2 + 3 + 4     ..................... 1000

if we sum all these we get 500500
Yeah, we simply had to sum the numbers 1 upto 1000
1000*1000-500*999
answered Jul 6, 2019 by (34,820 points) With the help of matrix rep of relation

1) find out direct all 1 entry.(in first solutions)

2)find out zero entry and than complement

1000*1000-500*999=500500

answered Jul 9, 2019 by (34,820 points)