5 boys and 5 girls are sitting in a row randomly. The probability that boys and girls sit alternately is...... ?

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There are total 10 seats. _ _ _ _ _ _ _ _ _ _. Case 1 1st seat is occupied by a boy. For this, the number of choices are 5. 2nd seat is occupied by a girl. For this number of choices again are 5 3rd seat is occupied by a boy. For this, the number of choices are 4. 4th seat is occupied...
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There are total 10 seats. _ _ _ _ _ _ _ _ _ _. Case 1 1st seat is occupied by a boy. For this, the number of choices are 5. 2nd seat is occupied by a girl. For this number of choices again are 5 3rd seat is occupied by a boy. For this, the number of choices are 4. 4th seat is occupied by a girl. For this number of choices again are 4 5th seat is occupied by a boy. For this, the number of choices are 3. 6th seat is occupied by a girl. For this number of choices again are 3 7th seat is occupied by a boy. For this, the number of choices are 2. 8th seat is occupied by a girl. For this number of choices again are 2 9th seat is occupied by a boy. For this, there is 1 choice only 10th seat is occupied by a girl. For this, there is 1 choice only. So total number of choices for Case 1 = 5*5*4*4*3*3*2*2*1*1 we can rearrange it and write this as = 5 *4*3*2*1*5*4*3*2*1 = 5!*5!. Case 2 Here we start with girl sitting in the first seat. Similar to the first case here also total ways of arranging the students alternately would be = 5!*5! Therefore total no. of ways = 2* 5!*5! Sample space [all arrangements] is = 10! as there are 10 seats. So, the probability of arranging them in alternate positions = 2* 5!*5! /10! by cancelling one 5! from numerator and denominator we get, = 2 * 5!/ (10*9*8*7*6) = 2 * 5*4*3*2*1/ (10*9*8*7*6) = 1/(9*2*7) = 1/126 read less
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After placing all the 5 boys, the girls can be placed in between them, and there are 6 places, as indicated by * as given below: *b*b*b*b*b* So, 5 boys can be arranged among themselves in 5! ways, and the girls can be arranged in possible 6 places in 6p5 ways=6!. So the total number of required arrangements...
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After placing all the 5 boys, the girls can be placed in between them, and there are 6 places, as indicated by * as given below: *b*b*b*b*b* So, 5 boys can be arranged among themselves in 5! ways, and the girls can be arranged in possible 6 places in 6p5 ways=6!. So the total number of required arrangements is 5!6!. Total number of possible arrangements (without any restriction) is 10!. So the required probability is 5!6!/10! = 1/42 read less
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Native English Speaker with 32 years experience in teaching English and communication skills.

5!6!/10! = 1/42
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GMAT Math Expert

there are two possibilities. One when boys are in odd places and other when boys are in even places. When boys are in odd places and girls in even places, each can be arranged in 5! ways. So 5! * 5! ways.. Again when boys in even places and girls in odd places, each can be arranged in 5! * 5! ways.....
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there are two possibilities. One when boys are in odd places and other when boys are in even places. When boys are in odd places and girls in even places, each can be arranged in 5! ways. So 5! * 5! ways.. Again when boys in even places and girls in odd places, each can be arranged in 5! * 5! ways.. So totally there are 2*(5! * 5!) ways read less
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Tutor - Maths and computer science

(2!.5!.5!)/ 10! = 1/126
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Headstart to assured 95+ score in math

(5! x 5! x 2)/ 10!
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5p5 = 120
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Spoken English, BTech Tuition, C Language, C++ Language, CAD etc., with 4 years of experience

(1- Probability that they are not sitting alternately)
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The number of ways to arrange boys is 5! ways : There are 6 places to arrange 5 girls. Selection 5 from 6 is 6C5 ways Probability = 5!x6C5 divided by 10!=120x6/10!= 1/10x9x8x7= 1/504
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http://gmatclub.com/forum/six-boys-and-six-girls-sit-in-a-row-at-random-find-the-57448.html
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