How many ways can you make the sum of a number

How many different ways can a number be expressed as a sum of any number of integers when order matters?

3 Answers 3

(Apply deductive/inductive reasoning) A nice number is a number that can be expressed as the sum of a string of two or more consecutive positive integers a) Determine which numbers from 50 to 70 inclusive are nice. Answer The numbers that are inclusive here are the n odd numbers from 1 to n = n2 just a matter of subtracting the squares. In this case, there are 5 odd numbers that are less than 10 like 10/5 which sums up to 52 = 25 or 102/4 going by this number or count input solution 1225 – 625 = 600 or (4900 – 2500)/4 = 600. The sum of the first n odd numbers is n2 considering that this is 502 = 2500 the first is square 1 tile making an arrangement of 2×2 the additional tile is contained on the right side which can be extended to an arrangement of 3×3 with an additional 2 tile on both sides with 1 at the corner right side thus 3+3+1 = 7 resulting to a series of 1+3+5+7+9+1. Let U denote the entire set of numbers. Let P and B denote the set of numbers from 50 to 70 inclusive are nice. Thus: |U| = 50 |B| = 70 |P| = 25 |P ∩ B| = 10

b) Show that numbers that are powers of 2 are nice Let the first set create the series of numbers that are of power 2 A= <… 1, 2, 4, 8, 16, 0.0625, 0.125, 0.25, 0.5 … >which is the same as A= <… 2^0, 2^1, 2^2, 2^3, 2^4, 2^(–4), 2^(–3), 2^(–2), 2^(–1) …>This value can be written as A= <2^n | n is N>where N represents the set of values that are infinite and countable with countable infinity. c) Show that positive integers which are not powers of 2 are not nice Suppose that 800×800= 640000 with the square of 800 ends containing four zeros with a positive integer square cannot end with four non zero equal digits with a known perfect square of 1111,2222,3333,…….9999 the values should be 5555 or 6666 or 9999 or 1111 or 4444 considering that the one’s digits squared number cannot be 2,3,7, or 8. d) Given a nice number, can you determine how many different ways it can be expressed as a sum of positive integers? n=3(1+2,2+1),n=4(1+3,3+1,2+2),n=5(1+4,4+1,2+3,3+2) n=2(1+1),n=3(1+2,2+1),n=4(1+3,3+1,2+2),n=5(1+4,4+1,2+3,3+2)n=2(1+1), Thus, the integral solution is x1+x2=nandx1,x2>0x1+x2=nandx1,x2>0 which is the same way used in finding the values of (n−2) without a 0 for either x1 or x2 which are the identical balls that fall under different distinct bins that places a cross divide of (n – 2) taking the values of divider across the values of (n – 2) being identical thus becoming.

How can I find the number of ways a number can be expressed as a sum of primes? [duplicate]

The number 7 can be expressed in 5 ways as a sum of primes:

Make a program that calculates, in how many ways number n can be expressed as a sum of primes. You can assume that n is a number between 0-100. Your program should print the answer in less than a second

Example 1:
Give number: 7 Result: 5

Example 2:
Give number: 20 Result: 732

Example 3:
Give number: 80 Result: 10343662267187

I’ve been at this problem for hours. I can’t figure out how to get n from (n-1). Here are the sums from the first 30 numbers by a tree search

I thought I had something with finding the biggest chain 7 = 5+2 and somehow using the knowledge that five can be written as 5, 3+2, 2+3, but somehow I need to account for the duplicate 2+3+2 replacement.

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Look up dynamic programming, specifically Wikipedia’s page and the examples there for the fibonacci sequence, and think about how you might be able to adapt that to your problem here.

There’re several options. Since you know the number is between 0-100, there is the obvious: cheat, simply make an array and fill in the numbers.

The other way would be a loop. You’d need all the primes under 100, because a number which is smaller than 100 can’t be expressed using the sum of a prime which is larger than 100. Eg. 99 can’t be expressed as the sum of 2 and any prime larger than 100.

Eg. 8 = 2 + 2 + 2 + 2, thus length of the sum is 4
9 = 2 + 2 + 2 + 3, thus length of the sum is 4

If you want performance you could cheat in another way by using GPGPU, which would significantly increase performance.

Then they’re is the shuffling method. If you know 7 = 2 + 2 + 3, you know 7 = 2 + 3 + 2. To do this you’d need a method of calculating the different possibilities of shuffling. You could store the combinations of possibilities or keep them in mind while writing your loop.

Here is a relative brute force method (in Java):

I understand this is complicated. I will try to use an analogy. Think of it as a combination lock. You know the maximum number of wheels, which you have to try, hence the «i» loop. Next you go through each possibility («j» loop) then you set the individual numbers («k» loop). The code in the «k» loop is used to go from the current possibility (value of j) to the actual numbers. After you entered all combinations for this amount of wheels, you calculate if any were correct and if so, you increase the number of possibilities.

I apologize in advance if I made any errors in the code.

In how many ways can I write n as a sum of integers?

I was wondering about the following combinatorial problem:
given a natural number n, for example 20, in how many ways can I write it as the sum of positive integers?
For example:
20 = 20
20 = 19 + 1
20 = 8 + 4 + 4 + 2 + 1 + 1
etc

Please note that I ignore the order of the numbers, i.e. 19 + 1 or 1 + 19 is one way, which makes that you can always write the largest number to the left, 2nd largest right of that one, etc.

Does there exist an explicit formula for this? I first thought of «the number of ways I can partition n non-distinguishable 1s in groups» which would be given (I think) by [tex]<2n-1 \choose n>[/tex], but this would give too much because it counts 19 + 1 as different from 1 + 19.

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But, that would be changing the rules. Negative numbers aren’t allowed.

I’ve done a lot of work studying integer partitions, and I’ve even started looking at a more complex case by making it N distinct objects instead of N parts. This changes the rule slightly. I’m still not interested in who holds which part of N, but now the parts are not only distinguished by size, but also by the contents.

I think that a lot of people are interested in integer partitions, because many people take the prime number challenge and they think that this may give them a clue. Aside from being able to calculate the probability of finding k eggs in an Easter egg hunt with N eggs, I really haven’t been able to find a practical application. This is why I haven’t devoted much time to looking at the partitioning of N distinct objects. I’ve been side tracked by something that I think is much more promising.

I remember this as a combinatorial problem:
Given [tex] n [/tex] identical cubes (or LEGO building blocks), in how many ways can we build [tex]k[/tex] towers with it. Example for [tex]n = 4 [/tex] and [tex]k = 2[/tex] (where we can do this in 2 ways)

The answer to this question is the following recursive formula (which does not has an algebraic solution as far as I know)

So the answer to your question would be

[tex]f(n, 1) + f(n, 2) + \dots + f(n, n)[/tex]

( why is there not a button to wrap input in a tex-tag)

How many ways can you roll a sum of 8 with two dice?

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.

So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be

P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

How many ways can you roll a sum of 8 with two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and
Sample space is
[ (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) ]

So, pairs with sum 8 are (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) i.e. total 5 pairs

So, in 5 ways we can roll a 8 with two dice.

Similar Questions

Question 1: How many ways can you roll a 7 with two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and
Sample space is
[ (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) ]

So, pairs with sum 7 are (1,6) (2, 5) (3,4) (4, 3) (5, 2) (6,1) i.e. total 6 pairs

So, in 6 ways we can roll a 7 with two dice.

Question 2: How many ways can you roll a 6 with two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and
Sample space is
[ (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) ]

So, pairs with sum 6 are (1,5) (2, 4) (3,3) (4, 2) (5, 1) i.e. total 5 pairs

So, in 5 ways we can roll a 6 with two dice.

Question 3: How many ways can you roll a 5 with two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and
Sample space is
[ (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) ]

So, pairs with sum 5 are (1,4) (2, 3) (3,2) (4, 1) i.e. total 4 pairs

So, in 4 ways we can roll a 5 with two dice.

Given a list of numbers, how many different ways can you add them together to get a sum S?

Given a list of numbers, how many different ways can you add them together to get a sum S?

list = [1, 2]

S = 5

This is what I’ve tried, but it only outputs 3 as the answer

However, this outputs three. I believe it outputs three because it doesn’t account for the different order you can add the numbers in. Any ideas on how to solve this.

The problem should work for much larger data: however, I give this simple example to give the general idea.

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Using both itertools.combinations_with_replacement and permutations :

How about using dynamic programming? I believe it’s more easy to understand and can be implemented easily.

The core idea here is using an extra record array to record how many ways can be found to get current num, e.g. record[2] = 2 means we can use to ways to get a sum of 2 ( 1+1 or 2 ).

And we have record[target] = sum(record[target-choices[i]]) where i iterates over choices. Try to think, the way of getting sum=5 must be related with the way of getting sum=4 and so on.

We suppose that your list consists of [1,2,5] so we have this recursive function :

so start from f(1) and work your way up with Dynamic programming. this solution in the worst case is O(n^2) and this happens when your list has O(n) items.

Solution would be something like this:

here is a working version.

You’d want to use recursion to traverse through each possibility for each stage of addition, and pass back the numbers used once you’ve reached a number that is equal to the expected.

You could write the last part a list comprehension but I think this way is clearer.

Combinations with the elements in a different order are considered to equivalent. For example, #3 and #5 from your list of summations are considered equivalent if you are only talking about combinations.

In contrast, permutations consider the two collections unique if they are comprised of the same elements in a different order.

To get the answer you are looking for you need to combine both concepts.