Calculating combinations?

Question

Is there a formula, for calculating the number of combinations for a certain outcome? If you could please, with all factors and element, give an explanation on how it works and if there is a formula. I've been working hard to look for certain patterns or trying to figure out if there's a formula (although I think it may be way beyond me and my classmates/peers in school) with my classmates, listing tables and etc.

An example for one:
We have three 6 sided dies
How many possible combinations are there for, say 11, with regards to the order of the combinations meaning 551 is different from 155 and 515 and 632 is different from 362 and 236 and so on and so forth.

Thanks in advance to anyone who can give me a good explanation and a formula with an explanation if there is a formula.

Answer 1

There are entire books on combinatorics where such formulas are given. It all depends on what you're trying to calculate.



Most fundamental is the function "n!' pronounced "n factorial". It's the product of all the integers leading up to that integer. For example, 5! = 5*4*3*2*1. The number n! tells you the amount of ways you can arrange n different objects in a straight line. This is because there are n different objects you can chose for the first one, n-1 for the second, n-2 left for the third, etc.



In this case we can just count. The three-number cominations that add up to 11 are (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3), and (4,4,3).



There are 3! different ways of having 6, 4, 1 show up. Ditto for 6,3,2 and 5,4,2. But for the others, we have a repeating digit, so there are only 3 different ways each of these can show up (e.g., 551, 515, 155).



So the number of cominations is 3!*3 + 3*3 = 18+9=27.

By the way, plural of "die" is "dice".

Answer 2

There are entire books on combinatorics where such formulas are given. It all depends on what you're trying to calculate.



Most fundamental is the function "n!' pronounced "n factorial". It's the product of all the integers leading up to that integer. For example, 5! = 5*4*3*2*1. The number n! tells you the amount of ways you can arrange n different objects in a straight line. This is because there are n different objects you can chose for the first one, n-1 for the second, n-2 left for the third, etc.



In this case we can just count. The three-number cominations that add up to 11 are (6,4,1), (6,3,2), (5,5,1), (5,4,2), (5,3,3), and (4,4,3).



There are 3! different ways of having 6, 4, 1 show up. Ditto for 6,3,2 and 5,4,2. But for the others, we have a repeating digit, so there are only 3 different ways each of these can show up (e.g., 551, 515, 155).



So the number of cominations is 3!*3 + 3*3 = 18+9=27.

By the way, plural of "die" is "dice".

Answer 3

For the sum of spots on a set of dice, probably the best method is simply careful counting, especially for as few dice as 3, where there will be only 216 such combinations regardless of any other conditions that must be met. This comes that there are 6 faces on which the first die can fall. For each one of these faces, there are 6 face on which the second die can fall. For each of these, the third die can fall on 6 faces. If you add more dice, you can generalize this to 6^n (6 to the nth power) where n is the number of dice.



Nevertheless, you can do some things to help speed up the counting. Let us say, for instance, that you want to count the dice in ascending order. Your possibilities are then



146

155

236

245

344

335



Each of these sets of n digits (n = 3 in this case) can be arranged in n!, or n * (n-1) * (n-2) * ... * 3 * 2 * 1 ways. For each set of unique digits, each permutation can be considered as a different number and all of the permutations can be counted. However, for every digit that is repeated k times, there are k! ways to arrange these digits, all of which reduce the total number of unique combinations of those digits by k!.



In this instance, three of the five combinations have all digits unique, giving 3! ways to arrange each one for a subtotal of 3 * 3! or 18 numbers whose digits add up to 11, plus three more with one repeated digit repeated twice. These two represent 3 * 3! / 2! or 9 more, for a total of 27 ways to roll 11 from three dice.

Answer 4

For the sum of spots on a set of dice, probably the best method is simply careful counting, especially for as few dice as 3, where there will be only 216 such combinations regardless of any other conditions that must be met. This comes that there are 6 faces on which the first die can fall. For each one of these faces, there are 6 face on which the second die can fall. For each of these, the third die can fall on 6 faces. If you add more dice, you can generalize this to 6^n (6 to the nth power) where n is the number of dice.



Nevertheless, you can do some things to help speed up the counting. Let us say, for instance, that you want to count the dice in ascending order. Your possibilities are then



146

155

236

245

344

335



Each of these sets of n digits (n = 3 in this case) can be arranged in n!, or n * (n-1) * (n-2) * ... * 3 * 2 * 1 ways. For each set of unique digits, each permutation can be considered as a different number and all of the permutations can be counted. However, for every digit that is repeated k times, there are k! ways to arrange these digits, all of which reduce the total number of unique combinations of those digits by k!.



In this instance, three of the five combinations have all digits unique, giving 3! ways to arrange each one for a subtotal of 3 * 3! or 18 numbers whose digits add up to 11, plus three more with one repeated digit repeated twice. These two represent 3 * 3! / 2! or 9 more, for a total of 27 ways to roll 11 from three dice.

Answer 5

combinations & permutations.



combinations nd permutations help you find the number of outcomes for an event.



with Permutations, order matters. 551 would NOT be the same as 155 or 515.



a permutation is what you're looking for.

nPr = n! / (n-r)!



Combinations, order doesn't matter.

nCr = n!/ (n-r)!r!



--examples.--

Suppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.

Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:





CA CT AC AT TC TA





---



When we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:



CA CT AT

We say '3 choose 2' and write 3_C_2.

Answer 6

The problem is equivalent to finding the number of solutions in positive integers not exceeding 6 of:



x1 + x2 + x3 = 11



The answer is 27.



For three (differentiated) 6-sided dice, if you want to achieve a total of N (3≤N≤18), the number of combinations is:



C(N-1, 2) - C(3,1)C(N-7,2) + C(3,2)C(N-13,2)



= C(N-1,2) - 3*C(N-7,2) + 3*C(N-13,2)

This is the formula you requested.



Here, C(a,b) is the combinitorial function

C(a,b) = a! / [(a-b)! * b!]

a! ('a' factorial) = a*(a-1)*(a-2)*...*3*2*1 ; 0! = 1

and C(a,b) = 0 if a


The proof of this formula is complex and rather difficult to explain (not to mention to type) here. For the workings of a similar problem,



see : https://answersrip.com/question/index?qid=20070411224523AAiowNx



The general solution to the number of solutions of:



x1 + x2 + x3 + ... + xk = m



in positive integers not exceeding the number 'c' (for 6-sided differentiated dice, c=6) is:



C(m-1, k-1)

- C(k, 1)*C(m-c-1, k-1)

+ C(k, 2)*C(m-2c-1, k-1)

- C(k, 3)*C(m-3c-1, k-1)

+ C(k, 4)*C(m-4c-1, k-1)

- ...



where the series continues until terms equalling zero occur.

Answer 7

Yes there is a formula. if you want to send me an email. My Email is [e-mail]