How many possibilities are there to group n IDENTICAL objects into k groups?
353352
The answer is Choose(n+k-1,k-1).\n\nFor example, if n=5 and k=3, the value is Choose(7,2) = 21.\n\nOne way to understand that is to take n+k-1 different pigeon holes:\n\n    O O O O O O O\n\nNow, if we choose 2 of these, say, the fourth and the sixth, like:\n    O O O X O X O\n\nthat yields the distribution as (3,1,1).  You can, I hope, see how this works in general.
674697
For five consecutive, there are 34 possibilities because you could start with 1, 2, 3,..., or 34\nYou can't start with 35 because then you would have\n35, 36, 37, 38 which is only four numbers.\nI'm not sure about the other two.\nNote: The answer to the other two are not 35 and 36.
975944
5 would be (a)\n6 would be (b)\n7 would be (c)\n8 would be (b)
138694
each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
1071317
There are exactly 343 combinations.  I can list them all, but others have already cluttered this thread quite well on their own.  ^.^\n\nYou can think of the solution set as a series of tiered combinations.  These would start with those sets beginning with a $50 bill, which can contain up to 2 x $50,  2 x $20, 5 x $10, etc. which I believe someone has started already in this thread.  Or group them as pairs, triplets, etc. with the equation 50j+20k+10l+5m+1n=100, which may be a bit easier for calculations.  Or a convergent series, where A=(ways to form $5), B=(ways to form $10, including A-sets), etc.  With any method, find the different combinations for each set, then add them all up.\n\nDoing this is still tedious and prone to mistakes, and thus probably not much of an improvement over manually listing all of the combinations one by one.  It can also quickly become an insane exercise for larger values (for example, add the $2 bill, or find change for $200, or even worse).  So instead of trying to think this problem through mathematically, I found it much easier to think it through as a programming problem.  It lends itself very readily to a simple recursive subroutine using the following pseudocode:\n\nmake an empty stack\nset total combinations to 0\ncall CheckCurrency\nprint total combinations found\n\nrecursive subroutine CheckCurrency\n- if total currency in stack = 100\n- - add 1 to the total combinations\n- - (optional: display stack contents)\n- - end recursion\n- if total currency in stack > 100\n- - end recursion\n- if total currency in stack < 100, then\n- - for each valid denomination (50,20,10,5,1)\n- - - if the denomination < = smallest denomination in the stack\n- - - - push it to the stack\n- - - - call CheckCurrency\n- - - - pop it from the stack\nend of subroutine\n\nThis required only a few minutes to design, another few minutes to code, another few minutes to debug, less than 2 seconds to run, and has the wonderful bonus of being able to display every single combination if desired.  It would also be trivial to modify it so that, for example, you can find all combinations if you also include the $2 bill.  (The answer is:  4562 combinations.)  Or to find all combinations totalling $200.  (The answer is: 2930 combinations.)\n\nThe line that checks if the denomination is less than or equal to the smallest denomination currently in the stack is present to avoid a common mistake:  We want the total combinations, not total permutations.  Otherwise, the stack "50 20 20 10" would be counted differently than the stacks "50 20 10 20" and "50 10 20 20", and even "10 20 20 50".  The total permutations is much, much greater, especially if you consider each bill as unique.  Five $20 dollar bills alone have 5!=120 permutations because, after all, they are all completely different $20 bills.  Imagine the number of permutations of 100 x $1 bills, and everything in between.  o.O
294686
The number of permutations of a string of n unique objects is n!. \nThis can be derived by the fact that if you have the set of all permutations of that string, all the elements of the set can be generated by first choosing an object to be first in the string, then a second object from the n-1 remaining objects to be second in the string, and so forth. By the fundamental principle of counting, one can thus generate n*(n-1)*(n-2)*...*1 = n! elements.\nIf some of the objects in the string are not unique, ie., the number of permutations of the string (E, E, E, V, R, S, T), we can show that the number of unique permutations is the same as the number of permutations given by pretending the objects are unique divided by the number of permutations of the non-unique objects.\nThis can be shown as follows: Suppose P(n) is the number of permutations of n unique objects. Suppose that P(n, r) is the number of permutations of n unique objects and r non-unique objects. Then by the fundamental principle of counting, if we replaced each non-unique object in the string from a set of r objects, we have P(n, r)*P(r) = P(n) permutations. Thus P(n, r) = P(n)/P(r) = n!/r!.
930557
Either we have BBBGGG or GGGBBB. In both cases, there are 3! ways of arranging boys, and 3! ways of arranging girls. So 3!*3!*2 = 72.
198636
The number of combinations is given by\nnCr  =  n! / ( (n-r)! r! )\n\nThis formula works by calculating the number of possible permutations of r=5 choices from n=38 objects then dividing it by the number of ways the 5 objects can be arranged leaving just the number of possible combinations without worrying about their order.\n\nn!/(n-r)! = 38 * 37 * 36 * 35 * 34 = 60233040\n\nThis is the number of different possible permutations.\n\nr! = 5 * 4 * 3 * 2 * 1 = 120\n\nThis is the number of ways you can rearrange 5 objects. Each different possible combination of 5 from 38 objects can be arranged in 120 different permutations. Thus taking the total number of permutations and dividing by 120 will tell you the number of unique combinations.\n\n501942
661462
Not likely to find this answer, as that any one can open hundreds of groups individually. Doesn't mean they are active and functioning, but there is an infinite amount of them that can be made.
1180538
Nope, that is the combination of {x,y,z} taking 2 objects at a time, because combination has no regard for specific order. The permutations for {x,y,z} taking two  at a time are: xy, xz, yx, yz, zx, zy.
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
638209
1, 2, 3, 6, 7, 14, 21, 42
905726
three sizes\nthree crusts\nten toppings\n\nLook in your math text under "combinations and permutations." You can do this!
943885
if you take the numbers one-by-one: 25! / ( (25-1)!*1! )\n\ncombination from n numbers taken by k:  n! / ( (n-k)!*k! )\n\nfor example:\n    -25 numbers, taken by 7: 25! / (18!*7!)\n    -25 numbers, taken by 4: 25!/ (21!*4!)
241816
It is certainly possible. I'll give you an example for H-O-Y\n\n\nFirst, let's find the index of the last group starting with H, which is H-Y-Z. Every group before H-Y-Z and H-Y-Z itself always includes a letter that comes before I, and every group after H-Y-Z are the only ones not containing the first 8 letters. Therefore, there are C(26-8,3)=816 groups. This makes H-Y-Z 2600-816 = 1784th group.\n\nNow, let's only consider the groups starting with H. Let's find the index of H-O-Z. Every group after H-O-Z are does not contain letters before P if we do not consider the first letter. There are C(26-15,2) = 55 such groups. Therefore, H-O-Z is 1784-55 = 1729th group.\n\nH-O-Y comes right before H-O-Z. Therefore, the index of H-O-Y is 1728.
348738
A partition of a positive integer n is a set of k strictly positive integers whose sum is n.\n\nHere are the eleven partitions of 6. Observe that they are given in reverse lexicographic order.\n\n{\n {6},\n {5,1},\n {4, 2},\n {4,1,1},\n {3,3},\n {3,2,1},\n {3,1,1,1},\n {2,2,2},\n {2,2,1,1},\n {2,1,1,1,1},\n {1, 1, 1, 1, 1, 1}\n}\n\n\nA set partition of n elements is a grouping of all the elements into nonempty, nonintersecting subsets.\n\nSet partitions are different than integer partitions, representing the ways we can partition distinct elements into subsets. They are useful for representing colorings and clusterings. \n\nSet Partitition of {1,2,3}:\n\n{\n {{1,2,3}},\n {{1}, {2,3}},\n {{1,2}, {3}},\n {{1,3}, {2}},\n {{1}, {2}, {3}}\n}
821077
once we chose 5 persons for one committe then other committe members get fixed.\n\nThe 5 members can be chosen in 10C5 ways = 10*9*8*7*6/120\nor 6*7*6 or 252 ways