In how many arrangements can 3 boys and 4 girls stand in a row such that no two boys are together?

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In how many arrangements can 3 boys and 4 girls stand in a row such that no two boys are together?

Mathematics
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Number of ways that they can stand in - number of ways two boys can stand together. Note that the latter also covers the number of ways in which three boys can stand together.
Oh, and also keep in mind that they can permute among themselves, like @ganeshie8 spotted the last time.

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I got 5040 at total number of ways they can stand together
How would you calculate the number of ways two boys can stand together? I feel like I am messing up there
Just like you did the last time.
Oh, and yes, you have three boys, so you also need to see that you can choose any 2 out of 3 boys.
See I messed up somewhere I got 5028...
4 girls can be arranged in 4! after that, 3 boys can be inserted in between the boys in 5C3 * 3! ways so in total we have 4! * 5C3*3! = 1440 ways
720 4,320 3,600 4,000
Why 5C3?
\[7! - 2 \cdot \binom{3}{2}\cdot 6!\]
place girls in the blanks first |dw:1432838542055:dw|
Oh, I see.
after that, we have 5 places for placing boys |dw:1432838582145:dw|
but 1440 is not an option. so what is wrong?
Is 720 correct? Just wondering...
720 is an option
Is the option correct?
I dont think so?
What's the correct one?
Im not sure :/
i don't see how it cannot be 1440
1440 is that arrangement is the only the girls are sitting on the end seats
Pattern: gbgbgbg # of arrangements: 4!*3! = 144
girls need not be in the ends |dw:1432839047302:dw|
# of ways to fill the end seats: 4*3 = 12 # of ways to fill the other 5 seats: 5! # of arrangements: 12*5! = 1440
yeah :/ I am getting the same way... but this is not an answer...
as the question stands, 1440 is the correct answer it just so happens that the options don't have the correct answer... inform ur teacher and move on to next q

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