The cards are shuffled and then arranged in a straight line.\nHow many arrangements are there where no two vowels are next to each other?
136154
farrell the first part of your solution is wrong.  There are 10 places to put the vowels, not 8, they can be at the beginning or the end. I believe it is 9! * (10!/5!) = 9! * 10 * 9 * 8 * 7 * 6\n\ntsniperx might be right but the question is ambiguous about this.  It says how many arrangements of the cards (not sequences of letters) are there which to me means that 0'0 and 00' are two different arrangements because the cards are in different order...but like I said it isn't obvious either way.  \n\nIF tsniperx is correct cornelius is still wrong, it would be 9!*10!/(5!*2!*2!*3!)  In other words, cornelius' answer is double the correct answer.  But I still think my original answer is correct.
118832
Ubergeek is right again, almost.  Half the cards are black, so prob that the first one is is 1/2.  After that, only 25 of the remaining 51 cards are, so prob that the second is black is 25/51, and prob both are is the product, 1/2 * 25/51 = 0.245098, roughly.  I used to teach this stuff.
357155
The probability is the number of desired outcomes divided by the total number of outcomes.\n\nThere are 4 vowels out of a total of 11 letters.  Each letter has equal probability so the chance of a vowel is 4/11 (or 36.3636%)\n\nSimilarly there are 7 consonants out of a total of 11 letters.  The probability of a consonant is 7/11 or 64.6464%\n\nFortunately these add up to 11/11 or 100%, so that confirms our answers.\n\nIn summary:\nThe chance of drawing a vowel is 4/11 or 36.3636%
357380
You are placing the letters MMMIINU into seven boxes, one per box.\n\nThe Ms can go in C(7,3) = 7!/(3!4!) = 35 ways.\n\nThere are four boxes left, and there are C(4,2)=6 ways to put choose two locations for the Is.\n\nThen there are two boxes left, and exactly 2 ways to choose where the N and U go.\n\nSo the total is 35*6*2=420.
554401
OOHHH! OHHH!!! I KNOW THE ANSWER TO THIS!!! ok, there is an equation for this, it's    n(n-1)/2.\n\nSo, if there are 100 "techies" then put 100 into the equation.\n\n100(100-1)/2------the equation\n100(99)/2------simplified\n9900/2------- 100*99\n4950-----THE ANSWER!!!\n\nSo, there would be 4950 cards exchanged!\n\nTHIS IS THE ANSWER, I PROMISE YOU!
507595
54 - if you count the jokers\n+12\n+190\n---------\n256\n----------\n\nor     52\n     + 13 - counting the number 12 as two seperate numbers\n     +190\n    ---------\n        2 5 5\n   ------------\n\nor 54+13+190= 257\n\nor Cards 52 + numbers on clock(counting each single digit) 15+ 190 = 257 with jokers in a pack of cards = 259\n\nso the answer is either 257, 259, 254, 256, 86 or 88\n\noooo! My head hurts!\n\n=kizzy=
126071
if the order you draw them is important its a permutation.\n\nthe answer is basically; a 1 in 55 chance for the first card, a 1 in 54 chance for the second card (you've taken the first card out) and a 1 in 53 chance of getting the final card. all three events are independant of eachother. so the answer is 55 x 54 x 53 = 157410\nso 1 in 157410 chance.\n\nif the order you draw them is unimportant them its a combination.\nfor the first card there is a 3 in 55 chance\nthe second card has a 2 in 54 chance\nand the final card has a 1 in 53 chance\nthe answer is: (3/55)  x (2/54) x (1/53) = 1 in 8745
53337
If you fold A in half (right onto left) the two halves match.  Same goes for H, I, M, O, T, U, V, W, X,and Y. If you fold B in half (top on to bottom) the two halves match.  Same goes for C, D, E, H, I, K, O, and X.  Instead of folding you could use a mirror along the centre line.
19522
52 cards in a pack with joker? maybe?
670801
The probability is 11/1105.\nYou mulitply all of the fractions- 12/52, 11/51, 10/50 and than simplify.\nBy the way, a standard deck has 52 cards, not 53 cards.
670951
There are 4 ducks in all, arranged this way:\n\n          x     x\n\n          x     x
520135
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
647018
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
538724
If you number the tpopping from 1 to 8, you can have the following combinations of different toppings (numbered for simplicity):\n1) 1, 2, 3\n2) 1, 2, 4\n3) 1, 2, 5\n4) 1, 2, 6\n5) 1, 2, 7\n6) 1, 2, 8\n7) 1, 3, 4\n8) 1, 3, 5\n9) 1, 3, 6\n10) 1, 3, 7\n11) 1, 3, 8\n12) 1, 4, 5\n13) 1, 4, 6\n14) 1, 4, 7\n15) 1, 4, 8\n16) 1, 5, 6\n17) 1, 5, 7\n18) 1, 5, 8\n19) 1, 6, 7\n20) 1, 6, 8\n21) 1, 7, 8\n22) 2, 3, 4\n23) 2, 3, 5\n24) 2, 3, 6\n25) 2, 3, 7\n26) 2, 3, 8\n27) 2, 4, 5\n28) 2, 4, 6\n29) 2, 4, 7\n30) 2, 4, 8\n31) 2, 5, 6\n32) 2, 5, 7\n33) 2, 5, 8\n34) 2, 6, 7\n35) 2, 6, 8\n36) 2, 7, 8\n37) 3, 4, 5\n38) 3, 4, 6\n39) 3, 4, 7\n40) 3, 4, 8\n41) 3, 5, 6\n42) 3, 5, 7\n43) 3, 5, 8\n44) 3, 6, 7\n45) 3, 6, 8\n46) 3, 7, 8\n47) 4, 5, 6\n48) 4, 5, 7\n49) 4, 5, 8\n50) 4, 6, 7\n51) 4, 6, 8\n52) 4, 7, 8\n53) 5, 6, 7\n54) 5, 6, 8\n55) 5, 7, 8\n56) 6, 7, 8\n\nThis means that there are 56 different topping combinations. Since there are 3 types of pizza, 168 (which is 56 * 3) is the number of different pizzas you can choose from here.\n\nAnswer: 168 different pizza type and topping posibilities.
648750
2 x 25!\n\nYou can have two kinds of palindromes ... one where the center letter repeats (e.g. NOON), and one where it doesn't. (NON).\n\nEach kind there are 25! permutations starting with the letter A.  so the total number of palindromes is 2 x 25!
115728
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
240516
Actually, the above answers are incorrect.\n\nWe have 5 cards and 4 suits.\n\nLet's, say we draw the cards in the following order:\nAce, 2,3,4,5\n\nThere are 4 suits. So there 4 ways we could draw an Ace first; 4 ways we could draw a 2, 2nd, etc. And so there are 4x4x4x4x4 = 1024 ways of drawing the cards in the above order. However the cards can be drawn in any order, so we have to multipy 1024 by the number of ways the five cards  could be ordered when they are drawn. The number of ways the 5 cards could be ordered is 5 factorial (i.e 5!) = 5x4x3x2x1=120. So the number of ways of drawing a hand consisting of the 5 specified cards, in any suit and in any order  =  1024x120 = 122,880