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March 9th, 2018, 01:11 PM  #1 
Newbie Joined: Jun 2017 From: Iraq Posts: 4 Thanks: 0  a question about permutations
Hi, There are 6 girls, 2 boys to be arranged in a row. How many ways N are there if The boys should be not together? please help 
March 9th, 2018, 03:18 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,647 Thanks: 955 
1 = girls, 2 = boys from 11111212 to 21211111 : 21 ways Or there is a total of 28 ways, of which the boys are together in 7. 
March 9th, 2018, 04:58 PM  #3 
Newbie Joined: Jun 2017 From: Iraq Posts: 4 Thanks: 0  
March 10th, 2018, 09:44 AM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 
I interpret your problem as having 8 different people of which 6 are girls and 2 are boys. That is the set $\{b_1,b_2,g_1,g_2,g_3,g_4,g_5,g_6\}$. So the number of permutations of the set with the b's together is $7\cdot2\cdot6!$ and the complement then has $8!14\cdot6!=42\cdot6!=30240$ 
March 10th, 2018, 12:52 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,647 Thanks: 955 
...and I took it as the 6 girls being identical sixtuplex and the boys identical twins!!


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