Can you help me with this problem on permutation please?
914386
The definition of permutation says n P r = n! / (n-r)!\nHere you have 7! / 0!\n7! = 7*6*5*4*3*2*1 = 5040\n0! = 1\nSo the answer is 5040 / 1 = 5040.
172875
Obviously you are doing homework and don't want to figure it out yourself.  It is easy, just write them out.  Obviously you can disregard the B, because there is no choice (both dominant).  Now do all the combinations with a "A" and then double your answer to get the "a".\nACD\nACd\nAcD\nAcd\nThat is four combinations.  Double it for "a" and you get \n\nEIGHT\n\nTyping out the question is harder once you learn the rules.
1161271
In all these problems you are going to pick a set of objects or perform a set of actions.  \n\nIf the ordering of the outcomes makes a difference, then it's a permutation. If the ordering doesn't make a difference, then it's a combination.\n\nExample 1: The 10 runners run a race. They give medals for the first 3 places. How many different outcomes are there?\n\nThe order of finish makes a difference, since it determines who gets which medal, so it's a permutation problem.\n\nAnswer: 10x9x8 = 720\n\nExample 2: Three students are chosen from a class of 10 to clean after a food fight. How many ways can the group be chosen?\n\nIt doesn't matter who gets picked first, they all have to clean. Ordering doesn't make a difference, so it's a combination problem.\n\nAnswer: 10C3 = 10!/3!7! = 10x9x8/3x2x1 = 120.
264022
This is the combination of 40 things taken 10 at a time :\n40 C 10 = 40! / (10!30!)= 847,660,528\n \nCombinations do not consider order or arrangement (i.e. a,b as against b,a).  When order is considered we take the permutation instead.
900956
If you include combinations with no people and all people, the answer is 32 ways.  2^5.\n\n{ } = 1\n{A}, {B}, {C}, {D}, {E} = 5\n{AB}, {AC}, {AD}, {AE}, {BC}, {BD}, {BE}, {CD}, {CE}, {DE} = 10\n{ABC}, {ABD}, {ABE}, {ACD}, {ACE}, {ADE}, {BCD}, {BCE}, {BDE}, {CDE} = 10\n{ABCD}, {ABCE}, {ABDE}, {ACDE}, {BCDE} = 5\n{ABCDE} = 1\n\nThat's a total of 32 ways.\n\nAnother way to look at is to think that each person can be included or excluded.  There are two ways to include/exclude person A, two ways to include/exclude person B, etc.  That's 2^5 = 32.\n\nThis all assumes that ordering doesn't matter (that a combination of person A and B is the same as person B and A).\n\n(Note: if no people in the set is considered invalid, just subtract 1 and get 31).
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
775189
the following is the total permutation of\nthe four numbers:-\n\n1234\n1243\n1324\n1342\n1423\n1432\n2134\n2143\n2314\n2341\n2413\n2431\n3124\n3142\n3214\n3241\n3412\n3421\n4123\n4132\n4231\n4213\n4321\n4312\n\ntotal amount of permutations=4!\n4*3*2*1=24\n\nnotice that 4 is grouped with 123 six times\nnotice that 3 is grouped with 124 six times\nnotice that 2 is grouped with 134 six times\nnotice that 1 is grouped with 234 six times\n\nthe six permutations of the three other\nnumbers =3!=3*2*1=6\n\nthere are two permutations of the other\ntwo numbers =2!=1*2\n\nhence,total for four numbers= 4*3*2*1\n=24\n\nthey are symmetric since each number\nhas to appear the same number of\ntimes in the permutation\n\ni hope that this helps
182985
No. of alphabets = 22  [{a, b, .. z) - {i, o, q, z)]\n==> n = 22\nNo. of letters taken at a time = 3 \n==> r = 3\n\nNo. of combinations of "n" different things taking "r" at a time is given by n_C_r or C(n,r) \n= (n)(n-1)..r times/(r) * (r-1) * ...(1)\n\nTherefore, the number of combinations that can be formed \n= C(22,3)\n= 22 * 21 * 20/3 * 2 * 1\n= 1,540\n\nIf you are looking for the number of words that can be formed using the letters,.. then you have to look out for permutations..\n[No. of permutations of "n" different things taking "r" at a time given by n_P_r or P(n,r) = (n) * (n -1) * ... r times.\n\nThe number of words that can be formed \n= P(22,3)\n= 22 * 21 * 20\n= 9,240\n\nIf the same letter can be used then the number of words that can be formed is given by (n)^r\nThe number words that can be formed \n= 22^3\n= 22 * 22 * 22\n= 10,648\n\n\nhttp://www.futureaccountant.com/probability/\n\nhttp://www.schoolingkids.com/
270029
There are of course good ways and bad ways of doing this.  The best way involves using a search tree like DFS and then pruning.  Instead of showing you something this complex (since I am unsure of your level of understanding) I will show you a common simple algorithm.  I will leave the code up to you, here is the psuedo-code:\n\n----------------------------------------------------\nALGORITHM JohnsonTrotter(n)\n//input: positive integer n\n//output: a list of all permutations of (1...n)\n\ninitialize the first permutation with (1,2...n) and with a second correspoing list ('L','L',...,'L').\n\n//Definition: mobile integer = if the corresponding letter is L \n//and the left adjacent number is smaller or vise versa.\n \nwhile there exists a mobile integer k do\n     find the largest integer K\n     swap K and the adjacent integer (left one if L) right otherwise\n     reverse the direction of all integers that are larger than k\nend do while\n----------------------------------------------------\nthis is really a simple algorithm to follow and to code, I am working off memmory here so it may not be the exact terms used but the algorithm is 100% I assure you.  Hope this helps and have fun.
438115
Okay, first you have to calculate how many the combinations of the first three teams are. At this point, arrangement is irrelevant. So, it'll be 15C3. Sorry, I don't exactly remember the full formula. Then, you'll get 455. Now, the arrangement is becoming relevant, since, the three teams could be number one, two, or three. So, at this point, permutation takes place. You are to multiply the previous answer with 3P3(or 3!, same things, aren't they?!). So,the answer will be 2,730! I know, it could be similar to another or other answer(s), but in math, the way of solving a question is more important than the answer itself! (That's what i think!)
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
109803
Combinations are basically what you do when order doesn't matter.For example, electing any 12 students to a group of 3. It would be 12 C 3.\n\n\nPermutations are what u do when order DOES mattter. For example, when your arranging 6 different books. The answer to it would be 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways.\n\nFactorials are basically. The number you have multiplied by everything less than that number.For instance, if you have 3. The factorial notation would be 3 x 2 x 1 = 6
631878
If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
303516
Each element x of S_n can be written as a product of disjoint cycles, e.g. x=(132)(45). Since the order of a cycle is the number of elements it permutes, the order of x is the least common multiple of the orders of each of its cycles.\nNow, the order of x should be 3000 = 2^3*3*5^3. To make the number n of elements permuted by x minimal, we have to find the set of coprime numbers d_i, i=1..k with d_1*...*d_k = 3000 that makes n=d_1+d_2+d_3 minimal. All possible choices are listed below :\n1. (k=3) = 3, 2^3, 5^3   \n2. (k=2) : 3*2^3, 5^3    \n3. (k=2) : 3, 2^3*5^3    \n4. (k=2) : 2^3, 3*5^3    \n5. (k=1) : 3*2^3*5^3     \nObviously, the first choice where k is maximal (= the number of different prime factors of 3000) minimizes n to the value 3+2^3+5^3=136.\n\njust in case you want to know... :\nQuite generally, the minimal n for an element x with order d is given by n = p_1^e_1+...+p_k^e_k where the p_i's are different prime numbers with d = p_1^e_1*...*p_k^e_k.
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).
191858
If there are no other constraints, the roles can be assigned in \n15*14*13*12*11*10 ways. The first role can be filled in 15 ways and the second in 14 ways and so on. This is also called a permutation. See http://mathworld.wolfram.com/Permutation.html
973961
a) i +(1/3)(j-i)\n\n  b)  \n1-12 =>1-4 & 5-12\n\n... 1-4=>1 & 3-4\n.......3-4=>3 & 4\n\n...5-12=> 5-6 & 7-12\n......5-6=> 5 & 6\n......7-12=>7-8 & 9-12\n..........7-8=>7 & 8\n..........9-12=> 9 & 10-12\n...............10-12=>10 & 11-12\n....................11-12=>11 & 12\n\n c) 1/3 of the computations lead to a subset of the previous set\nwhich is 1/2 the size of the other subset which is lead to twice\nas often. So it averages out to the same number of\ncomputations as a conventional binary search.