Discrete Math, Permutation question, how can I solve this answer?
929249
How do you know the answer is 96?  Wouldn't you need to use the P(n,r), or 6 choose 3?  \n\nThe formula (if that is correct) would be n!/(n-r)!  \n\n6x5x4x3x2 divided by (6-3)!.  This would equal 120.\n\nBut I don't know.  I'm taking a damn discrete math course myself and we're working on the same junk!  good luck!
247655
0,1,1,5,7,11,13,13,21,23,27.......
220111
Greetings:\n\nHaving selected one of the four, the probability that person #2 will have a different date of birth is 364/365 (if we seek no repeats, #2 can have any 364 of 365 possible b.days).  Similarly, having selected two, the probability that #3 will have a unique d.o.b. is 363/365 (any 363 of 365 possible dates okay).  Finally, for #4, the probability is 362/365.  Hence by the multiplication property, the probability that all four will have different b.days is given by, (364/365)(363/365)(362/365) = (364)(363)(362)/(365)^3 = 0.984 to three places.\n\nRegards,\n\nNikkor180
944835
Yes, you're right.  The only way to get a parralel line w/ x=6 is another vertical line.  So if x=4, the only possible perp line is y=6. Always hated those "trick" questions!  They seem so hard, yet are ridiculously easy.
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
531502
12408 = (2^3)*3*11*47\n\nso just pick all the combinations:\n1, 2, 4, 6, 8, 22, 24, 33, 44, 66, 88, 94, 132, 141, 264, 282, 517, 1551, 3102, 6204, 12408...there are probably more
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
561865
A permutation. P(n,r)=(n!)/((n-r)!). C(n,r)=(n!)/(r!(n-r)!)\nor (P(n,r))/(r!). x/something is smaller than x itself,(unless there's negatives or decimals, but we don't have to worry about that because r is positive) so the permutation is biggger than the combination.
724915
simple.  if you want to find the permutations of something of a set of numbers (or letters or objects), as long as you want to find just how many combinations there would be, you use factorial (!).  8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1...and that's your answer.
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
652572
This is a case of Circular Permutations where the formula is given by (n-1)! , taking into account clockwise and anti-clockwise arrangements as separate ones.\n\nSo your answer is 10 ! = 3628800 \n\nIf you are not taking into account clockwise and anti-clockwise ..then its 10 ! /2
676828
Done it:\n\nA = 3\n\nB = 9\n\nc = 6\n\nyou're welcome, if you have doubts of how i did it... erminsho@yahoo.com.mx
1103566
# combinations of 3 elements out of 5 = # combinations of 2 elements out of 5 = 5*4/2 = 10\n\n5 ways to pick one zero\n4 ways to pick another zero\n2=2! permutations per pair of zeros.
789066
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
747450
Sounds like Combinatorics.  Combinations, permutations, a bit of graph theory, discrete math, I usually describe it as "Advanced Counting".
242452
This is a typical combination problem.\n\nformula for combination is n!/r!(n-r)!\n\n= 9! / 5!(9-5)!\n= 9! / 5!4!\n= 126 (ANS.)\n\nfactorial ex. 5! = 5 x 4 x 3 x 2 x 1\n\nDONT FORGET TO RATE THIS AS BEST ANSWER
990622
if you know permutation/combination..the answer is easy to understand...\n\nlike 23C23+23C22+23C21+......+23C1