(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:

\(252 - 56 = 196\)