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Marriage of Integers: Disregard other thread

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by VP_Jim » Wed May 13, 2009 10:57 am
First, let's assume that B=10. That means that A is either 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, there are 9 different pairs we can make if B=10 (e.g., B=10 and A=9, or B=10 and A=8, etc).

What if B=9? Then, A must be 1, 2, 3, 4, 5, 6, 7, or 8. Thus, there are 8 possible pairs if B=9.

If B=8, there are 7 possible pairs. If B=7, there are 6 possible pairs, etc. All the way down to B=2, in which case A=1, so there is only 1 possible pair if B=2.

Therefore, the answer is:

9+8+7+6+5+4+3+2+1 = 45.

Hope this helps!
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by sureshbala » Wed May 13, 2009 12:13 pm
If I choose any two numbers from 1 to 10, definitely one of them will be less than the other.

Hence the total number of pairs = 10C2 = 45
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by KICKGMATASS123 » Thu May 14, 2009 10:53 am
VP_Jim wrote:First, let's assume that B=10. That means that A is either 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, there are 9 different pairs we can make if B=10 (e.g., B=10 and A=9, or B=10 and A=8, etc).

What if B=9? Then, A must be 1, 2, 3, 4, 5, 6, 7, or 8. Thus, there are 8 possible pairs if B=9.

If B=8, there are 7 possible pairs. If B=7, there are 6 possible pairs, etc. All the way down to B=2, in which case A=1, so there is only 1 possible pair if B=2.

Therefore, the answer is:

9+8+7+6+5+4+3+2+1 = 45.

Hope this helps!
Why Can't A = 10 when B = 10??
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by VP_Jim » Thu May 14, 2009 1:16 pm
Because the problem says that a<b... NOT a =< b. Therefore, if B=10, A must be less than 10.
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Re: Marriage of Integers: Disregard other thread

by dtweah » Thu May 14, 2009 3:43 pm
dtweah wrote:There are how many pairs of positive integers (a,b) such that 1 <=a < b <=10?
(a) 24
(b) 36
(c) 45
(d) 64
(e) 81


c) For a = j, j = 1,2,…,9 there are 10 – j pairs. Thus the answer is 9 + 8 + 7+ … + 1 = 45
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Re: Marriage of Integers: Disregard other thread

by KICKGMATASS123 » Thu May 14, 2009 5:42 pm
Thanks guys for clarifying.. I wasn't reading :(
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