Where can i find a plan to cover 6 numbers from 15 selections with best chance of repeating three numbers of 6
370301
The best way of covering all 15 numbers evenly is simply to start by selecting 1,2,3,4,5,6 then 7,8,9,10,11,12 then 13,14,15,1,2,3 etc. That way, you're sure to hit each number an equal amount of times in the long run. If you're saying "but these combinations are totally unlikely to be selected!" then yes, you are right. But they are precisely just as likely as any other of the combinations. So, in other words, don't play the lottery.
217328
pick yo own number 34,6,16,7,15 good luck
960222
here's the results for today's draw:\nhttp://www.myit.com.my/fdresult/default.aspx
66293
can you repeat numbers?\nif not\nfor the first digit you get 9 choices\nfor the second one 8\nfor the third 7\nand the forth 6\n\nso the total number is 9*8*7*6= 3024 combination\n\nif you can repeat numbers 9^4=6561\n\n(if you could use the zero, and were allowed to repeat numbers the answer would be 10^4=10000;that numbers from 0001 to 9999 and 0000, a total of 10,000 numbers)
283919
Find some 6 x 24 graph paper, or take two pieces of lined notebook paper and place them so they form squares and place them on a light box (or window).  You may need to open the curtains first.\n\nHold your 6 x 24 graph so that the 6 part runs along the bottom.  That is, hold your graph paper in "portrait mode".\nthat is like this: |, not like this __.\n\nNow label the X axis on your graph paper.  You need to know that the next whole number after -1 is 0, and then 1 and so on.  This part labels the vertical lines on the graph, and the numbers run horizontally.\n\nNow label the Y axis on your graph paper.  You need to apply the same funny rule for adding 1 to a negative number to know that the next number after -4 is -3, and so forth.  These labels are for the horizontal lines, and the numbers run vertical next to each other.\n\nYou now have a graph that labels 6 * 24 points, from (-1,-4) to (5,20).  Now you are ready to plot some of the points:\n\ny = 3x.  Substitute 0 for X.  y then = 3*0 = 0.  Thus (0,0) is a point to plot.  Place a big black spot on (0,0).\n\ny = 3x.  Substitute 1 for X.  y then = 3*1 = 3.  Thus (3,1) is a point to plot.  Place a big black spot on (3,1).\n\nRepeat this for the infinitely many solutions where y=3x fits on your graph.  You may be able to fool your teacher that you have done it infinitely many times by noticing a pattern that the spots fall on and FAKE the rest.
835896
if you have a set of numbers such as..... 1, 4, 5, 6, 5, 7, 5, 3, 8\n5 is the mode because it appears the most times in the set of numbers
1150319
Go to a Staples, Office Max or Office Depot. They all sell a variety of planners in different shapes, sizes and such.
895078
go to:\nwww.whateverlife.com\nthey have the best codes for any type of layout... ive been using them for months now!
182799
Sas will do it for you.\n\nHere is the article:\nhttp://support.sas.com/techsup/technote/ts498.html\n\nIf you know visual basic (and you can do this from excel using VBA) here is some code for generating combinations:\n\nhttp://www.codeguru.com/vb/gen/vb_misc/algorithms/article.php/c5607/
216068
Each cover can be associated with a stamp.. in 3 different ways...\nTherefore the 6 covers can be combined with the 3 types of stamps in 3 * 3 * 3 * 3 * 3 * 3\n= 3^6\n= 729\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
789443
http://www.bplans.com/sp/businessplans.cfm - you can use the formats on here and keep the words down to 1500 words - not sure you'll find one except someone will do it specifically for you!!
889463
A pattern in numbers.  i.e. - 1,1,2,3,5,8,13,etc (add each number to previous number to get next number)
764788
Here is a place I've had good luck with.\n\nhttp://baldwincooke.com/Default.aspx?FocusId=ProductDetail.ascx&ItemNumber=BC541&sendroicid=97c6c568-76ed-40f5-a82f-ff03b69b95d4&sendroikwd=monthly+planner
1027584
make one no... but you can go to freecodesource.com and they have some premade layouts... don't know if you'll find what you're looking for though
754620
you can only get 6 - 36.you can only get 6 one way, and 36 one way.  you get 7 six ways, and 35 six ways.  figure out all the ways to get each number, including 19, and add them all up.  the odds of getting 19 is the number of ways to get 19 over the total number of outcomes
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.
182775
there are 6 place for 15 student so:-\n\n15*14*13*12*11*10=answer\n\nor simply use combination since the arrangement is not important\n\n15C6