I have integer for example "123", using this i want to create all possible combinations listed below. 123 12,3 1,23 and so on irrespective of the digits i have entered. Is there any way possible using python for the same? I am not able to get any idea.
2272
I have a list with 15 numbers in, and I need to write some code that produces all 32,768 combinations of those numbers.   I've found  (by Googling) that apparently does what I'm looking for, but I found the code fairly opaque and am wary of using it. Plus I have a feeling there must be a more elegant solution.  The only thing that occurs to me would be to just loop through the decimal integers 1–32768 and convert those to binary, and use the binary representation as a filter to pick out the appropriate numbers.   Does anyone know of a better way? Using map(), maybe?
86190
Let T = {1000,1001...,9999}. How many numbers have at least one digit that is 0, at least one that is 1 and at least one that is 2. For example, 1072 and 2101 are two such numbers.  I have no idea to solve such probability question, I should use Permutation or Combination? why the answer is 150.  Thanks you guys.
6646
Background: I'm working on  for , and I've run into a problem with how I'm creating the individual matchups (groups of 4 entries).  Currently, I'm simply generating all combinations of 4 elements (in the combinatoric sense) of entries, but that gets really big really quickly:  $$x \choose 4$$    (from )  Therefore, my question is: How can I choose subsets (of length $n$) of a set such that each element of that set appears in the same subset as every other element at least once, while maintaining a minimal amount of subsets?  I can figure this out for $n = 2$, since the solution is quite intuitive: simply combine the first element with every other element, combine the second element with every other element that comes after it, etc. For a set of length $L$, this yields:  $$\sum_{x=1}^{L-1}(L - x)$$    (from )  which is much smaller and more manageable.  However, I still have not been able to figure out: How can I generalize this to any $n$?
75196
I am trying to find a way for the positive integers written as the sum of other positive integers.( expressed in terms of some functions)  I searched a bit and I came across with  But in my case the order matters. What I mean is for example if we write for 4,  there would be 7 ways to express 4 as;           1 + 3           2 + 2           3 + 1           1 + 1 + 2           1 + 2 + 1            2 + 1 + 1           1 + 1 + 1 + 1   Can anyone guide me in approaching to the solution of my problem? Thanks in advance.
28318
I'm trying to format numbers. Examples:  1     =&gt; 1 12    =&gt; 12 123   =&gt; 123 1234  =&gt; 1,234 12345 =&gt; 12,345   It strikes as a fairly common thing to do but I can't figure out which filter I'm supposed to use.  Edit: If you've a generic Python way to do this, I'm happy adding a formatted field in my model.
95227
I have a question and got strucked on this.. To find the total no. of six digit numbers that can be formed having property that every succeeding digit is greater than preceding digit. Please guide me to solve this. Thanks in advance.
120061
Shouldn't be repeated in any case; if the numbers appeared is exactly the same as another but different orders, that doesn't count as a new set. For example: $11,13,18,19,19,20$ is the same as $13, 18, 19,11,19,20$.
16994
I know how to generate a random number in PHP but lets say I want a random number between 1-10 but I want more 3,4,5's then 8,9,10's. How is this possible? I would post what I have tried but honestly, I don't even know where to start.
40270
I need to calculate no of possible substrings containing "00" as a substring. I know the length of the binary string.  Eg: for a string of length 4, possible substrings are: 0000 0001 0010 0011 0100 1000 1001 1100  I just need the number of possible combinations, not enumerate all of them.
81118
How many 6 digit numbers are possible with at most three digits repeated?  My attempt:  The possibilities are:   A)(3,2,1) One set of three repeated digit, another set of two repeated digit and another digit (Like, 353325, 126161)  B)(3,1,1,1) One set of three repeated digit, and three different digits.(Like 446394, 888764)  C)(2,2,1,1) Two sets of two repeated digits and two different digits (Like, 363615, 445598)  D)(2,2,2) Three sets of two repeated digits (Like, 223344, 547547)  E)(2,1,1,1,1,1) One set of two repeated digit and four different digits (Like 317653, 770986)  F)(1,1,1,1,1,1) Six Different digits (like 457326, 912568)  G)(3,3) Two pairs of three repeated digits. Let's try to calculate each possibilities separately.    F) is the easiest calculate.   Let us try to workout  Case E)  Let's divide the case into two parts:  Case E(1) Zero is not one of the digit   We can choose any $5$ numbers form $9$ numbers $(1,2,3,\cdot, 9)$ in $\binom{9}{5}$ ways , the digit which one is repeated can be chosen in 5 ways, and you can permute the digits in $\frac{6!}{2!}$ ways. The total number of ways$=\binom{9}{5}\times 5\times \frac{6!}{2!} $  Case E(2) Zero is one of the digit.  Case E(2)(a) Zero is the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times  (\frac{6!}{2!} -5!)$.  Case E(2)(b) Zero is not the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the repeated digit can be chosen in 4 ways,  the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times 4\times  (\frac{6!}{2!} -5!)$.  Before, proceeding to workout the other cases, I want to know   Is my attempt correct? If it is correct, it is too lengthy, is there any other way to solve this?
99761
How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3?  $$$$ I was trying to to add 2^6 (when there is no 2)+ C(6,2)2^2 (when 2 can be in two places)+C(6,3)2^3 (when 2 can be in three places)...
4701
How do I generate a random int value in a specific range? I have tried the following, but those do not work: Attempt 1: randomNum = minimum + (int)(Math.random() * maximum);  Bug: randomNum can be bigger than maximum. Attempt 2: Random rn = new Random(); int n = maximum - minimum + 1; int i = rn.nextInt() % n; randomNum =  minimum + i;  Bug: randomNum can be smaller than minimum.
64710
I can form numbers with only 0,2,4,6,8.  The sequence is as follows  0,2,4,6,8,20,22,24,26,28,40,42,.....  How to generate an ordered sequence of numbers from combinations of 0,2,4,6,8.
116996
What is the number of ways we can choose three letters from $\{A,B,...,Z\}$ if repetition is allowed?      This is not a permutation.  Is there any formula to solve this kind of problems?
107093
I've been asked the following question:  What three odd integers from the set {1,3,5,7,9,11,13,15} that when summed together equals to 30? Note that any integer can be used more than once.  Is there any possibility to solve these kind of questions with some formulae? Note I have gone through the answers for making it with 5 numbers.
80374
If I have this string:     2+24*48/32   what is the most efficient approach for creating this list:     ['2', '+', '24', '*', '48', '/', '32']
12867
How many $N$ digits binary numbers can be formed where $0$ is not repeated. Note - first digit can be $0$.  I am more interested on the thought process to solve such problems, and not just the answer. If anyone can cite some resources for learning how to solve such problems would be great.