permutations and combinations?
359078
C(4,1)C(4,1)C(4,1)/C(52,3) = 64/22100 = .0028959\n\nb)\nC(4,1)C(4,1)C(4,1)/52^3 = 64/140608 = .00045517
636772
k=1/4 or k=2 or many more possibilities
83369
I am giving you a method which would minimize number of possibilities.\nFactorize the number of factors( in your example 9) in as many way as possible.( in your example 9 and 3*3)\nNow find out the possible numbers ( one for one factor combination) in following way ( we are going to have 2 numbers)\nThen all factor combinations are sorted so that largest one is in first place and lowest one in last place.\nFor a factor combination x1*x2*x3*...xn*...\nthe number will be \n2^(x1-1) * 3^(x2-1) * 5^(x3-1) * ... * pn(xn-1)...\nwhere pn is nth prime number.\nFor our two combinations,\nfor 9-> 2^(9-1)=256\nfor 3*3-> 2^(3-1) * 3^(3-1) = 36\nLowest of these numbers is our result. \nIn our example it is 36.\nLet us see another example lowest number with 12 factors.\n12 factorizes to 12, 4*3, 6*2, 3*2*2\nfor 12-> 2^(12-1)=2048\nfor 4*3-> 2^(4-1) * 3^(3-1) = 72\nfor 6*2-> 2^(6-1) * 3^(2-1) = 96\nfor 3*2*2-> 2^(3-1) * 3^(2-1) * 5^(2-1)=60.\nSo our answer is 60.\nThis method will be if the number of factor cobination  is too large. An way to get good answer( may not be correct) is to start from prime factorization and continue untill collecting first two terms into one causes better result.\n\nThat is in case of 12\nWe will start from 2*2*3 and get 60.\nThen test if 4*3 is better. If not we will take 60 as answer.( as it is 72 so we will).\nThis case we have correct result but modified version may not produce correct result.
532000
How about 9! 51, 52, 55, 56, 57, 58, 59, 60, 61
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.
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
427494
The number of combinations is determined by multiplying the number of possible numbers of the first by the number of possible numbers of the second, third and last. If the any of the numbers could be used more than once, the answer would be 4 X 4 X 4 X 4 = 256. Since this problem says that the number can only be used once, then the number of possible choices for the first is 4, the second is only 3, the third is 2, and the last is only 1. So the way to arrive at 24 is 4 X 3 X 2 X 1. Maybe an example to make sure it is clear. If the first digit was 3, then the second could only be one of 5, 8, or 9. If the second was chosen as 5, then the third could only be one of 8 or 9. The last then could only be the one not selected as the third.
237350
Do you mean permutations?\n8P4\n= 8!/(8-4)!\n=8!/4!\n= 8*7*6*5\n= 1680
262
Permutation\nSuppose 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.\n\nBecause 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:\n\n      CA   CT   AC   AT   TC   TA\n\nCombination\nWhen 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:\n\n      CA   CT   AT
588383
Since to=o^5t, every element of <t,o> can be written in the form o^nt^m. Since t has order 2 and o has order 12, this gives 24 elements possible. Since H has elements that are odd permutations, H intersect A7 will have index 2 in H, so the collection of even permutations in H will have 12 elements as will the collection of odd permutations.
20923
the formula of permutation of 2 items is n(n-1)\nthe formula of combinations of 2 items is 0.5n(n-1)\nso u see easily that the first is the double of the second
499304
For each of the 4 positions in the password, there are 26+26+10-3=59 different choices.  Since there are no other restrictions, the number of permutations is:\n\n59x59x59x59=12,117,361
905726
three sizes\nthree crusts\nten toppings\n\nLook in your math text under "combinations and permutations." You can do this!
110013
Well, all numbers would have to be represented equally, right?  If there are 27 combination I assume that 1, 1, 1 is included in these combintations, not just every combination that includes all 3 numbers...  However, if all three numbers are represented equally, each number should appear 27 times.  So the answer should be 27*9, or 243.
801584
If the number begins with 5, there are still three of the odd digits available to fill the last place.  If a 3 is used for the last place, the others are filled using 1, 2, 3, 4, and so there are 4! arrangements.\nIf 1 is used for the last place, that leaves 2, 3, 3, 4 for the others, and so the number of arrangements is  4!/2! because of the repeated 3.  So that gives \n4! + 4!/2! = 36\n\nIf it begins with a 3, the last place can be filled with 1, 3, or 5, and  the numbers left to fill the remaining four places are all different and so the number of possibilities is\n3*4! =72\n\nIf it begins with a 2 or 4 ( two choices), and the last with 1 or 5 (two choices), there are 4!/2! ways of filling the other places because 3 is repeated.\n2*2*4!/2! = 48\n\nIf it begins with 2 or 4 and ends with 3, then the other four digits are all different, and the number of possibilities is\n2*4! = 48\n\nAdding all these together gives 204.  Yes.
18114
This is true only if the items are distinct. For example if the data set is the 26 letter alphabet and we take letters without replacement, then the number of permutations (order matters, so a,b is different from b,a) is 26*25 while the number of combinations (order does not matter, taking a, then b is the same as taking b, then a) is 26*25/2. \n\nIf our data set contains copies, like the letters in this sentence, then combinations of identical letters would not count doubly (so, a,a is not a seperate permutation from a,a).
342259
6! or 720 arrangements.\n\n2 * 4! or 48 with vowels at the ends, so it's a 1/15 chance.\n\nEdit: Yeah, I messed up.  Forgot the *2.