# In how many of the arrangements in a row of all ten bricks in Question 12 are: (a) the three bricks separated from each other (b) just 2 of the red bricks next to each other

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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).