How many unlabelled groups of 10 people can be formed?
There is a group of 500 people , each are from 6 different country and with 3 different profession. How many groups of 10 people can be formed with below condition. 1) no individual should belong to more than one group. 2)every group should have at least one person from every country. 3)every group should have at least 2 people from each profession.
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How to create all possible combinations of these two lists?
I'd like to take from these two lists to create a list of all combinations, where each combination is also a list.
E.g.
Given two lists:
[1,2,3]
and[True, False]
Combinations:
[(1, False), (2, False), (3, False)] [(1, False), (2, False), (3, True )] [(1, False), (2, True ), (3, False)] [(1, True ), (2, False), (3, False)] [(1, False), (2, True ), (3, True )] [(1, True ), (2, False), (3, True )] [(1, True ), (2, True ), (3, False)] [(1, True ), (2, True ), (3, True )]
There should be
2^n
combinations wheren
is the number of numbers.EDIT:
Tried to do the following:
[(n, b)  n < [1,2,3], b < [True, False]] (,) <$> [1,2,3] <*> [True, False]

Calculating all possibilities for a string with optional parts
I want to generate a list of all possible combinations of a string that contains optional parts. This is probably best explained with some examples:
A[B]
→A
andAB
A[B][C]
→A
,AB
,AC
andABC
A[B[C]]
→A
,AB
andABC
I hope this sufficiently explains what I'm trying to do.
I could hack together my own little parser or "algorithm" for this, but I have a strong feeling that there's an existing (and easier) solution for this. Because I don't have any kind of CS education (yet), I have no idea what kind of algorithm I'm looking for or even just what search terms to use.
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I have below one problem regarding permutation and combination. I know one solution which I am providing here. But I have another approach to the same problem but it is not giving me same answer as previous one. Can someone tell where am I making mistake here.
Problem: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done?
First Answer: We can select 5 men ...(option 1) Number of ways to do this = 7C5 We can select 4 men and 1 woman ...(option 2) Number of ways to do this = 7C4 × 6C1 We can select 3 men and 2 women ...(option 3) Number of ways to do this = 7C3 × 6C2 Total number of ways = 7C5 + (7C4 × 6C1) + (7C3 × 6C2) = 756.
Below is my new approach, where I am making mistake but not able to understand it.
atleast 3 men should be there. So ways to choose 3 men out of 7 = 7C3 = 35. Now 2 person has to be selected from remaining 4 men and 6 women. The no of ways it can be done = 10C2 = 45. Therefore, total no of way = 35*45 = 1575.
Can someone tell me what I am missing in second approach.
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