Question 758671: there are 5 boys and 3 girls .in how many ways can they stand in a row so that no girls are together
Found 4 solutions by lynnlo, 55305, ikleyn, math_tutor2020:
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Answer by math_tutor2020(3828) (Show Source):
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Answer: 14400
Explanation
There are 5 boys and 5! = 5*4*3*2*1 = 120 ways to arrange the boys.
Let's give the boys names such as:
Alex
Bill
Carl
Dave
Eddy
Or simply A,B,C,D,E for short.
Place blank markers between each pair of adjacent letters.
Also, place blanks at the start and end of the sequence.
We'll have this
_, A, _, B, _, C, _, D, _, E, _
Each blank space represents a possible spot for a girl to be placed.
This is to guarantee the girls are not together (i.e there aren't any girls paired up nor are the 3 girls in one block).
There are 5 boys and 5+1 = 6 blank slots.
Use the nCr combination formula to determine there are 6C3 = 20 ways to select the 3 blank slots.
Once a blank slot trio is selected, there would be 3! = 3*2*1 = 6 ways to arrange the girls in those slots.
To recap quickly:
120 ways to arrange the boys
20 ways to pick the 3 blank slots
6 ways to arrange the girls for any given slot trio selection
Therefore we have 120*20*6 = 14400 different arrangements where the girls are not together, i.e. no girls are next to each other.
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