Lining up 10 people in a line, and 2 groups of 2 must be togehter, how many ways can you arrange that line?
545283
there are two groups of 2 people , so take those 2 people as 1 unit. so there are total 8 units.\nthus number of ways to be arranged is 8! , that is 40320.
578450
What question are you asking? They're in a line and they would have to happen all at once to make a difference.
821077
once we chose 5 persons for one committe then other committe members get fixed.\n\nThe 5 members can be chosen in 10C5 ways = 10*9*8*7*6/120\nor 6*7*6 or 252 ways
859995
well, it will depend on the size"width" of the people, as to how many will fit,,,, you could try taking the adverage width of the people you have, using that as a base, then factor in your width of the space, and also if the people are going to overlap  or stand shoulder to shoulder to get your width,,,,, then you would have to figure you adverage  "depth " of the people,  \n\nfor example, if your people are 2 ft wide at their widest point, and you space is 10 ft wide, you can estimate you can fit 5 people across,, if you space is also 10 ft deep, and your people are 1foot deep at their widest depth, you can fit them in 10 deep,,you can fit 5wide x 10 deep ,,, so 50 people could go into a 10ft by 10 ft space,   your variables would be the difference (how much each varies from your estimate)  the larger your group/space   the more your estimate would adverage out accurate,,,,,,,,, i would think the only way to know , is to literally do it, have the actual people in that space, or their actual mesuraments,,,, as i said, the other variable is if you overlap their shoulders,,,
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).
233027
There are 8! different arrangements (permutations) of the 8 friends.\n\nThere are 4! different arrangements of the 4 boys.\n\nThere are 5 different locations where the leftmost boy can sit when the 4 boys sit together.\n\nTherefore, there are 5*4! (or 5!) arrangements of the boys where all the boys sit together.  But there are also 4! arrangements of the 4 girls in the remaining slots.  So there are 5!*4! arrangements where the 4 boys sit together.\n\nSince there are 8! arrangements overall, the probability of all boys sitting together is (5!*4!)/8!.\n\nSimilarly, if the boys sit to the left and the girls sit to the right, then the probability is (4!*4!)/8!.\n\nInterestingly, this last answer is also the same probability as boy-girl-boy-girl-boy-girl-boy-girl and also the probability of girl-boy-girl-boy-girl-boy-girl-boy.  Therefore, the probability of alternating seats is (2*4!*4!)/8!.\n\nRecall that n! (read n-factorial) is n*(n-1)*(n-2)*...*3*2*1.
392841
fill 7 and tip into 10\nfill 7 again and tip into ten leaving 4\ntip out ten and put 4 into 10\nfill 7 and tip into 10 leaving 1\ntip out 10 and pour in 1\nfill 7 again and pour into 10 leaving 8
30239
You can't just draw one? It's 10 boxes across the top and down the side, starting with 0 and going to 9 (and no, the corner doesn't share both sides). All you need are a pen and a ruler.
927347
Draw lines from the top of the vertical lines in the H to the opposite lower corners. There you go.\n\nIn response to the comment below:\nBTW\nI guess the middle line of the H would have to be not directly centered between the lines.  It has to be a little highter or lower so the 2 lines do not intersect on the center line.  Then you come out with 5 individual triangles and not triangles within triangles.
460943
You can't do it in three.  You need four lines to complete.  Click below for the solution.
7441
give them like 10 bucks if there'e 2-3 people, 7 if there's one
940694
Im assuming that by using the word 'groups', (plural) you are not allowed all four in one group as a solution.\nSo, if the counters are the same, then 2 ways:\n1, 1, 1, 1 \nand\n2, 2\n\nIf the counters are different, then 4 ways.  (If the counters are:\na, b, c, d then:\nab,cd\nac,bd\nad, bc\na, b, c, d. )\n\nIf this is for a 3rd grader then the former is probably what they are looking for, but I can't really know that.  The latter seems a little complicated for the average 3rd grader.
1040385
1/(52!)\nThere's only one correct order.\nThe number of ways to arrange 52 objects is 52!
303088
Tear a piece of paper up into ten pieces.  Write down an option on each and then fold them up.  Now, mix them up a bit and then draw one out.  Do whatever option you draw.  Have Fun!!!
246807
I can't count, so don't blame me if I get this wrong.\nLet n be the number of participants.\nSince it's a close circuit, the people in front of him are in a sense also behind him so there ara (n-1) people both ahead and behind him.\nso (1/5+5/6)(n-1)=n\n31/30*n -31/30=n\n1/30*n=31/30\nthus n=31\n31 total people
68596
1) keep "people" the same; balance "minutes" \noperation: solve this proportion\n\nprevious sheets : NEW SHEETS = previous minutes : new minutes\n\nNEW SHEETS = previous sheets * new minutes / previous minutes\n\nNEW SHEETS = 2 * 10 / 5 = 2*2 = 4\n\nat this point you know that 2 people can work 4 sheets in 10 minutes.\n\nNow keep minutes the same. Balance sheets with a proportion:\n\nNEW PEOPLE : old people = new sheets : old sheets\n\nNEW PEOPLE = old people * new sheets / old sheets\n\nNEW PEOPLE = 2 * 20 / 4 = 2 * 5 = 10\n\n10 is the answer
765309
It depends if one person can be in a room or if all 4 people can be in the same room.  If only one person can be in a room at any given time, then it would be \n4*3*2*1=24 different ways\n\nIf everyone can go in the same room, then there are 4^4=256 different ways.