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March 9th, 2018, 12:11 PM   #1
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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
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March 9th, 2018, 02:18 PM   #2
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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.
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March 9th, 2018, 03:58 PM   #3
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but the answer of lectures is
30240!


the following solution is the full example with answer ...

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March 10th, 2018, 08:44 AM   #4
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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 $8!-14\cdot6!=42\cdot6!=30240$
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March 10th, 2018, 11:52 AM   #5
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...and I took it as the 6 girls being identical sixtuplex and the boys identical twins!!
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