Set Q is defined as odd integers from 1 to 2000. How many ordered pairs (a,b) are there in set Q whereby a<b?

for any odd number (b), there are (b-1)/2 odd numbers less than it.

check on that thinking. for b= 11, the odd numbers less are 1,3,5,7,9 or five. Yep, it works.

There are 1000 numbers for the choice of b, so there must be this number of ordered pairs:

Sum b=1 to 1999(0dd) of (b-1)/2

Sum 1/2 (b-1)= 1/2 sum b - 1/2 sum
and see if you can sum those from 1 to 1999

Thanks for responding, it really helped.

To find the number of ordered pairs (a,b) in set Q where a<b, we first need to determine the number of odd integers from 1 to 2000.

In this case, since we have a range of numbers, we can calculate the number of odd integers using the formula:

Number of odd integers = (Last number - First number)/2 + 1

First number = 1
Last number = 2000

Number of odd integers = (2000 - 1)/2 + 1
Number of odd integers = 1999/2 + 1
Number of odd integers = 999 + 1
Number of odd integers = 1000

So there are 1000 odd integers from 1 to 2000.

Now, to find the number of ordered pairs (a,b) where a<b, we can consider the possibilities for a and b. Since a<b, we have:

For a = 1, there are 999 possible values for b (2, 3, 4, ..., 1000).
For a = 2, there are 998 possible values for b (3, 4, 5, ..., 1000).
For a = 3, there are 997 possible values for b (4, 5, 6, ..., 1000).
And so on...

We can observe that the number of possible values for b decreases by 1 as a increases.

Therefore, we can calculate the total number of ordered pairs (a,b) where a<b by summing the number of possible values for b for each value of a:

Total number of ordered pairs = 999 + 998 + 997 + ... + 1

Using the formula for the sum of an arithmetic series, we get:

Total number of ordered pairs = (Number of terms / 2) * (First term + Last term)
Total number of ordered pairs = (999 / 2) * (1 + 999)
Total number of ordered pairs = 999 * 500
Total number of ordered pairs = 499,500

Therefore, there are 499,500 ordered pairs (a,b) in set Q where a<b.