a)
The total number of ways
of arranging 10 bricks in a row = 10P10 =10!.
The number of
ways the 3 bricks staying together consecutively in all arrangements is as good as
treating those 3 bricks as a one single block for pemuting purpose.This block together
with other 7 bricks, we have to arrange in 8 places in 8places . This is possible
in 8P8 = 8!. But within the block 3 paticular bricks, they could be arranged in 3! ways.
Thus the number of arrangements of particular blocks are together = 3!*8P8 =
3!*8!
Therefore 10! - (3!*8!) is the number of ways of
arrangements where particular 3 blocks are not
together.
ii)
Particular 2 red
bricks are next to each other- with this condtion we treat the two particular red bricks
as one block and the remaing 8 bricks as different. So we arrange the 9 different things
on a row. This is possible in 9P9 = 9!. If the two red bricks are also distinct between
themselves, then they could be arranged 2! ways between themseves within the block. Then
the arrangments become 9!*2!
Therefore the number of
arrangements that 2particular red bricks are together = 9! (red are not distinct between
themselves). Or 9!*2! (if reds are distinct between themselves).
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