#### Topic: Sexual segregation

7 men and 8 women randomly sit down to round desktop with 15 chairs. What probability of that that on two standing chairs casually selected nearby 2 women will sit?

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7 men and 8 women randomly sit down to round desktop with 15 chairs. What probability of that that on two standing chairs casually selected nearby 2 women will sit?

Hello, qwp, you wrote: qwp> 7 men and 8 women randomly sit down To round desktop with 15 chairs. qwp> what probability of that that on two standing chairs casually selected nearby 2 women will sit? Like here without a dirty trick all: 8/15 * 7/14 = 0.266 (6)

Hello, andy1618, you wrote: A> Hello, qwp, you wrote: qwp>> 7 men and 8 women randomly sit down To round desktop with 15 chairs. qwp>> what probability of that that on two standing chairs casually selected nearby 2 women will sit? A> like here without a dirty trick all: 8/15 * 7/14 = 0.266 (6) At me such layout Variants on two nearby standing chairs of MM 7/15*7/15 7/15*8/15 7/15*8/15 8/15*8/15 Required probability the Total of probabilities turned out - 1 Your variant implies that after on one chair the woman sat down, we still had 14 chairs and 7 women. But we select at once two chairs....

Hello, Titus, you wrote: T> Hello, andy1618, you wrote: A>> Hello, qwp, you wrote: qwp>>> 7 men and 8 women randomly sit down To round desktop with 15 chairs. qwp>>> what probability of that that on two standing chairs casually selected nearby 2 women will sit? A>> like here without a dirty trick all: 8/15 * 7/14 = 0.266 (6) T> At me such layout T> Variants on two nearby standing chairs T> MM 7/15*7/15 T> 7/15*8/15 T> 7/15*8/15 T> 8/15*8/15 Required probability T> the Total of probabilities turned out - 1 T> Your variant implies that after on one chair the woman sat down, we still had 14 chairs and 7 women. T> but we select at once two chairs.... On your logic if we select 15 chairs, that is probability that on all there will be women that cannot be.

Hello, Qulac, you wrote: Q> Hello, Titus, you wrote: T>> Hello, andy1618, you wrote: A>>> Hello, qwp, you wrote: qwp>>>> 7 men and 8 women randomly sit down To round desktop with 15 chairs. qwp>>>> what probability of that that on two standing chairs casually selected nearby 2 women will sit? A>>> like here without a dirty trick all: 8/15 * 7/14 = 0.266 (6) T>> At me such layout T>> Variants on two nearby standing chairs T>> MM 7/15*7/15 T>> 7/15*8/15 T>> 7/15*8/15 T>> 8/15*8/15 Required probability T>> the Total of probabilities turned out - 1 T>> Your variant implies that after on one chair the woman sat down, we still had 14 chairs and 7 women. T>> but we select at once two chairs.... Q> On your logic if we select 15 chairs, that is probability that on all there will be women that cannot be. Not truly. When we select 15 chairs, the variant *15 will not be present among the possible. Another matter, whether among all possible variants the total of probabilities will be equal 1 (at such logic)? To tell the truth, I do not know to apply even the formula about number possible ... And as the logic andy1618 too gives 1, there can be you and MM 7/15*6/14 0.2 7/15*8/14 0.266666667 7/14*8/15 0.266666667 8/15*7/14 0.266666667 Total 1 Even are right, I think, what yes, andy1618 the rights

Hello, Qulac, you wrote: Q> On your logic if we select 15 chairs, that is probability that on all there will be women that cannot be. By the way, I will be grateful, if specify the formula for calculation of variants in all 15 chairs. Really 15!?

Hello, Titus, you wrote: T> Hello, Qulac, you wrote: Q>> On your logic if we select 15 chairs, that is probability that on all there will be women that cannot be. T> by the way, I will be grateful, if specify the formula for calculation of variants in all 15 chairs. Really 15!? Not, not similar. It would be valid, if each of men differed from another, but on statements of the problem they are faceless.

Hello, Titus, you wrote: T> By the way, I will be grateful, if specify the formula for calculation of variants in all 15 chairs. Really 15!? See about Necklaces in combinatorics

Hello, the Corkcrew, you wrote: Hello, Titus, you wrote: T>> By the way, I will be grateful, if specify the formula for calculation of variants in all 15 chairs. Really 15!? See about Necklaces in combinatorics I Think that it is not necessary. "The necklace" task is one of classical combinatorial tasks. It is required to count an amount of various necklaces from n , each of which can be painted one of k colors. At comparing of two necklaces they can be turned, but not to overturn (i.e. it is authorized to make ring shift). The necklace, yes is just hints at a dirty trick in desktop. The formula 2^15 (the maximum number of combinations in 15 number) does not consider shift possibility, and the necklace considers. And here restriction: exactly seven it is colored in one color and exactly eight in other color, - not registration.

Hello, Titus, you wrote: T> Your variant implies that after on one chair the woman sat down, we still had 14 chairs and 7 women. T> but we select at once two chairs.... And ? In you the task is easier: there are two chairs, on one peaks fine-molded, on other, 1 man and 1 woman, probability of what them will sit two men, whether is equal 1/2*1/2 that?

Hello, Titus, you wrote: T> the Necklace, yes is just hints at a dirty trick in desktop. The formula 2^15 (the maximum number of combinations in 15 number) does not consider shift possibility, and the necklace considers. T> and here restriction: exactly seven it is colored in one color and exactly eight in other color, - not registration. Restriction "exactly seven is colored in one color and exactly eight in other color" is considered in the formula of an amount of combinations. It is enough to us to seat men of 7 persons on 15 chairs, on the remaining women sit down. An amount of combinations 7 from 15 = 6435. This amount generally. And turns are considered in Lemme Bernsajda mentioned under the link. At us receives such "necklace" that whatever one may do it, it is impossible to receive coincidence of colors before turn, except for "zero" turn. At "zero" turn an amount of stabilizators equally 6435, at remaining turns equally 0. Amount of various turns equally 15. Therefore under the formula total , taking into account turns, i.e. the amount of orbits, will be equal (6435+0+0 +... +0)/15 = 429. It is possible to look for example here more in detail: the Lemma of Byornsajda of the Orbit

Hello, andy1618, you wrote: A> Hello, qwp, you wrote: qwp>> 7 men and 8 women randomly sit down To round desktop with 15 chairs. qwp>> what probability of that that on two standing chairs casually selected nearby 2 women will sit? A> like here without a dirty trick all: 8/15 * 7/14 = 0.266 (6) It turns out that round desktop . does not add? And the task is reduced in two ases from a pack?

Hello, Kodt, you wrote: And it is more than that, "nearby standing" too does not add .. It was not retained, : private double RoundDeskProbability () {const int tossCount = 100000, male = 1, female = 2; int ok = 0; Random r = new Random (); int [] chairs = {1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2}; for (int i = 1; i <tossCount; i ++) {//in a random way we transplant people for desktop int [] rand_chairs = chairs. OrderBy (az => r. Next ()).ToArray (); int ch1 = r. Next (0,14); int ch2 = (ch1 == 14? 1:ch1+1); if (rand_chairs [ch1] == 2 && rand_chairs [ch2] == 2) ok ++;} return (double) ok/tossCount;} the Theory converges with practice

Hello, Titus, you wrote: T> By the way, I will be grateful, if specify the formula for calculation of variants in all 15 chairs. Really 15!? Obviously "Tse on 8 from 15" it and on seven, by the way... In general 15!/7!/8!

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