# Combinations of Numbers

by Raabi Anony
(Karachi, Pakistan)

Hello everybody

It is my very first posting and hope receiving help and guidance from the Math Geeks. Here is my very 1st problem:

I have 2 equal sets of numbers 1 to 6 as below:

A = {1, 2, 3, 4, 5, 6}
B = {1, 2, 3, 4, 5, 6}

How many unique pairs of numbers can be made out of these 2 sets, like;
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
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(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

While manually doing, I get total number of 36 pairs and after deducting the repetitions, I get 32 pairs. But, by using the Combination Formula 12C2 (12 numbers taken 2 at a time), I get 66 pairs. Where am I wrong!
Actually, I want to solve this problem for a larger number of sets, like 10. I will appreciate clearing my confusions and suggesting the better algorithm, if any.

Please note that I am interested in the number of pairs only, just for a general purpose, without getting into any definition of Sets or the Ordered Pairs etc.

Thanks in anticipation.

Raabi

### Comments for Combinations of Numbers

Average Rating     Apr 08, 2013 Rating     Actually... by: Verona Your question is about permutations, not combinations. That is, in your problem rolling a 3 and a 2 is different from rolling a 2 and a 3. While this is not usually the case in dice rolling problem, you can say that you have an orange die and a purple die, in order to distinguish them. After here i will use "n" to mean "number". Anyways, the first step is to find the n of possible rolls on each die: 6. same on the other die. Lets say you've rolled a 2 on the orange die. Once this has been done, there are 6 outcomes: {6,1} {6,2} {6,3} and so on until {6,6}. Now we multiply this n by the n of possibilities on the orange die, because 6 is the n of outcomes when the first die has already been rolled. 6 * 6 = 36 Hope this helped! I challenge you find the number of outcomes without repetition, and when order of rolling does not matter (ie the n of combinations)

 Apr 15, 2012 Rating     The same question in another way by: Raabi Anony Sorry, I want to put my question in another more explanatory way. Let's say, there are 2 dices. When I roll these 2 dices, I get the following pairs: 1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) The above pairs of numbers equal 36. But, when I use the the Combination Formula 12C2 (12 numbers taken 2 at a time), I get 66 pairs. Naturally, the discrepancy is the result of my confusion, which I want to understand.