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Feb/Mar 2020 p52 q4
2768
Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line.
(a) Find the number of different arrangements of the 11 candles if there is a red candle at each end.
(b) Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together.
Solution
(a) With a red candle at each end, we have the arrangement R _ _ _ _ _ _ _ _ _ R. This leaves 9 positions to fill with 3 blue and 6 green candles. The number of arrangements is given by:
\(\frac{9!}{3!6!} = 84\)
(b) Treat the 3 blue candles as a single block, so we have B B B as one unit. This leaves us with 7 units to arrange: B B B, G, G, G, G, G, G, and 2 red candles. First, arrange the 7 units:
\(\frac{7!}{6!} = 7\)
Now, consider the arrangements where the red candles are not together. Calculate the total arrangements of the 7 units and subtract the arrangements where the red candles are together:
Total arrangements with blues together: \(\frac{9!}{6!} = 84\)
Arrangements with blues and reds together: \(\frac{8!}{6!} = 56\)
So, the number of arrangements where the red candles are not together is:
