Figuring out # of possible combos of 3

  • Thread starter Math Is Hard
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In summary, a friend asked for help in figuring out how many unordered combinations of three can be made from 15 items. The formula for this is {n\choose k}={n!\over k!(n-k)!}, which in this case would be {15!\over 3!12!}, resulting in 455 possible combinations. The formula is derived by considering the number of ways to choose each letter in the combination, taking into account duplicates, and dividing by the number of ways the letters can be rearranged.
  • #1
Math Is Hard
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A friend asked me this question, and I am not sure I know how to answer it:


************************************************
I have 15 items, and I want to figure out how many unordered combinations of three there are. Any ideas?

a b c d e f g h i j k l m n o

Example:
abc bcd cde
abd bce cdf
abe bcf cde
abf bcg cdg

************************************************


The way I think it can be done is using this formula:
[tex]
{n\choose k}={n!\over k!(n-k)!}
[/tex]

so you would have [tex]{15!\over 3!12!}[/tex] or 455 possible combinations.
But I am not really confident in this because I never studied this formula in a class - and I don't know if I am using it correctly.

Thanks!
-MIH
 
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  • #2
Think about how you would choose such combinations. Start by choosing the first letter: there are 15 ways to do that. Now choose the second letter: 14 ways. Finally, choose the third letter: 13 ways. There are 15*14*13 ways. You could write that as 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1/12*11*10*9*8*7*6*5*4*3*2*1= 15!/12!= 15!/(15-3)!

But that counts "abc" and "bac" as separate which you don't want to do. For each 3 letters there are 3! ways to get those same letters in different orders. To account for that, divide the answer above by 3!. That gives [tex]{15!\over 3!12!}[/tex]
exactly as you suggest. The same analysis for k objects chosen from n gives [tex]{n\choose k}[/tex] (which, in tex form is written "n choose k"!).
 
  • #3
Thanks so much, Halls!
I appreciate you taking the time to show me how that works.
 
  • #4
Hey does anyone know physics?
 
  • #5
Sharayah, if you have a physics question, post in under General Physics, and create a New Thread.
 

What is the formula for calculating the number of possible combinations of 3 items?

The formula for calculating the number of possible combinations of 3 items is n! / r!(n-r)!, where n is the total number of items and r is the number of items being chosen for each combination. In this case, n=3 and r=3, so the formula becomes 3! / 3!(3-3)! = 3!/3!0! = 3.

How does order affect the number of possible combinations?

The order in which the items are chosen does not affect the number of possible combinations. For example, the combinations ABC and CBA are considered the same because they contain the same items. This means that the number of possible combinations remains the same regardless of the order in which the items are chosen.

What is the difference between combinations and permutations?

Combinations and permutations both involve selecting items from a larger set, but they differ in terms of order. Combinations are a selection of items without regard to order, while permutations are a selection of items with a specific order. For example, the combination ABC is the same as the permutation ABC, but the permutation ABC is different from ACB.

What is the difference between combinations and variations?

Combinations and variations are similar in that they both involve selecting items from a larger set, but they differ in terms of repetition. Combinations do not allow repetition, meaning that once an item is chosen, it cannot be chosen again. Variations, on the other hand, do allow repetition, meaning that an item can be chosen multiple times. For example, the combination ABC is the same as the variation ABC, but the variation AAB is different from the combination AAB.

Can the formula for calculating combinations be applied to larger sets of items?

Yes, the formula for calculating combinations can be applied to larger sets of items. The only change is the values of n and r in the formula. For example, if there are 10 items and we want to calculate the number of combinations of 5 items, the formula becomes 10! / 5!(10-5)! = 10!/5!5! = 252.

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