<PRE><B>Question 1108562</B>
The 3 that must be together can arrange
themselves 6 different ways:
a b c
a c b
b a c
b c a
c a b
c b a
P( 3,1 ) = 3! / 1!
P( 3,1 ) = 6 
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Each one of these 6 possible groupings can 
fit into the arrangements of the other 3 students:
( think of the 3 that must be together as one
student that 6 different identities )
{{{ 6*P( 4,1 ) }}}
{{{ 6*( 4!/1! ) }}}
{{{ 6*4*3*2*1 = 144 }}}
144 different ways
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Definitely get a 2nd opinion if needed
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