in how many ways you can divide 4 coins between 3 persons(a,b,c)provided-a- gets one?
448949
If a gets one coil, there are four ways.\n\na(1), b(3), c(0)\na(1), b(2), c(1)\na(1), b(1), c(2)\na(1), b(0), c(3)
892661
8 numbers can be placed  in 8! ways.\nExchange the three two's gives no new numbers, so divide by 3!\nExchange the two six's gives no new numbers, so divide by 2!\n8! / ( 3! x 2! ) = 1680\n\nTh
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.
954059
there is a logical way to indicate solutions\n\nfind smallest A such that 1/ A < 3/7\nA= 4\nNow if A = 4\n1/B+1/C = 3/7 -1/4 = 5/28\n\n5/28 = (1+4)/28 = 1/28+1/7 \n            (2+3)/28 no\nif we take A = 5 we may get more solutions\n\n1/B+1/C= 3/7-1/5= 8/35\n= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no\nno other\n\ntake A =6 and we get\n\n1/B+1/C = 3/7-1/6 = 11/42\nno other solution as 11/42 is not sum of reciprocal
577048
depends you take away 3 and give it to whom? if to u keep it to urself then you have 5...if you eat the 3 then you have two...if you take the three away and leave the two you have 3...now im confused!
709555
There are 4*3*5*4*3*4*3*2=17280 possible account numbers.
353352
The answer is Choose(n+k-1,k-1).\n\nFor example, if n=5 and k=3, the value is Choose(7,2) = 21.\n\nOne way to understand that is to take n+k-1 different pigeon holes:\n\n    O O O O O O O\n\nNow, if we choose 2 of these, say, the fourth and the sixth, like:\n    O O O X O X O\n\nthat yields the distribution as (3,1,1).  You can, I hope, see how this works in general.
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
540886
According to the question the answer is 3
719127
Well you got a quarter and a nickel.  Because one of the coins is not a nickel, it is a quarter.  So the other one is a nickel.  Hehehe.... Good one.
816805
i doubt u got it right. this is how u do it:\n\nfirst u group them \n\n(x³-3x²)+(4x-12)\n\nthen simplify each binomial\n\nx²(x-3)+4(x-3)\n\nthen since (x-3) are the same on both of them u group them together\n\n(x²+4)(x-3)\n\nand there u have it
882485
1 Nickel = 5 c\n1 Dime = 10 c\n1 Quarter = 25 c\n5.50$ = 550 c\n\nAssume number of nickels = X\n\nThen number of quarters = X + 4\n\nThen number of dimes = total number of coins – (number of nickels + number of quarters) = 37 – (X + X + 4) = 33 – 2X\n\nnumber of dimes * dime value + number of nickels * nickel value + number of quarters * quarter value = Total value\n\n(33 – 2X) * 10 + X * 5 + (X + 4) * 25 = 550\n330 – 20X + 5X + 25X + 100 = 550\n10X + 430 = 550\n10X = 550 – 430\n10X = 120\nX = 120 / 10\nX = 12\n\nNumber of nickels = X = 12\nNumber of dimes = 33 – 2X = 33 – 2 * 12 = 9\nNumber of quarters = X + 4 = 16\n\n12 * 5c + 9 * 10c + 16 * 25c = 60c + 90c + 400c = 550 c = 5.50 $\n\n12 + 9 + 16 = 37 coins\n\nAnswer: Jo has 9 dimes :-)
860554
27 ways to pick the first person in the group\n26 ways to pick the second\n25 ways to pick the third\n24 ways to pick the fourth\n23 ways to pick the fifth.\n\nHowever, order is irrelevant, so we need to divide by 5!.\n\nThis is 27 choose 5\n= 27 x 26 x 25 x 24 x 23 / 120 \n= 80,730 ways
388148
All seniors with at least 2 girls means either 2 girls and 2 boy or 3 girls and 1 boy or 4 girls and 0 boys.  There are not 4 senior girls so the last cannot occur.\nC(3,2)C(3,2) = 3*3 = 9\nC(3,3)C(3,1) = 1*3 = 3\n9 + 3 = 12\nThe total possible ways of picking 4 people is\nC(12,4) = 495\n12/495 = .024242\nb)\nmore girls than boys means 3 girls and 1 boy or 4 girls and 0 boys\nC(7,3)(C(5,1) = 35*5 = 175\nC(7,4) = 35\n175+35 = 210\n210/435 = .424242
259226
here's one with coins that was on one of the worksheets i have:\n\nlet's say, for example, that you have $8.75 in dimes and quarters and there are 65 coins. how many dimes and quarters are there?\n\nso there would be 2 equations:\nd+q=65 (which is how many coins there are)\n\nand\n\n10d+25q=875 (how much money there is in dimes and quarters)\n\nfrom there, you'd solve it up and down like\n\n 10d+25q=875\n+  d+  q= 65 (Multiply by 10)\n---------------------------\n 10d+25q  =875\n+(-10d-10q=-650)\n-----------------------\n       15q= 225\n         q=15\n\nthen you plug in 15 for q for one of the equations.\n\n15+d=65\n65-15=50\nd=50\n\nyou'd have 50 dimes and 15 quarters. (i hope)\n\ni also hoped this helped ^^;
622584
A coin is flipped three times.\n\nA is the event that there is a head on the first flip.\nB is the event that there is a head on the second flip.\nC is the event that there is exactly one head in three flips.\n\nP(A) = 0.5\nP(B) = 0.5\nP(C) = 3/8 = 0.375 (If you flip a coin three times, there are 8 possible outcomes; of those, three result in exactly one head. You can also use the binomial here.)\nP(A|C) = P(A and C)/P(C)\n= (1/8)/(3/8) (Because only HTT will allow A and C to both happen, which is one outcome out of eight.)\n= 1/3\nP(B|C) = 1/3 (Same reasoning as above.)\n\nSo A and B are independent, but are not independent of C.\n\nP(A and B|C) = P(A and B and C)/P(C)\n\nNow if A and B, happens, then the first two flips were heads, meaning that there are at least two heads. But if C happens, there can only be one head, so\n\nP(A and B and C) = 0\n\nSo,\n\nP(A and B|C) = 0\n\nHowever P(A|C)*P(B|C) = (1/3)^2.\n\nSo I guess that the equality doesn't hold. And that also means that there is no positive constant K that will make it hold since K would have to be zero in this case.
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