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. Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_m20_qp_52 Year：2020 Question No：4(b)
Answer:
\(\wedge(\mathrm{BBB})^{\wedge} \wedge \wedge \wedge \wedge\)
\[
\frac{7 !}{6 !} \times \frac{8 \times 7}{2}
\]
\[
=196
\]
Alternative for question 4(b)
[Arrangements, blues together-Arrangements with blues together and reds together \(=\) ]
\[
\begin{array}{l}
\frac{9 !}{2 ! 6 !}-\frac{8 !}{6 !}
=[252-56]
\end{array}
\]
\[
=196
\]
\[
\frac{7 !}{6 !} \times \frac{8 \times 7}{2}
\]
\[
=196
\]
Alternative for question 4(b)
[Arrangements, blues together-Arrangements with blues together and reds together \(=\) ]
\[
\begin{array}{l}
\frac{9 !}{2 ! 6 !}-\frac{8 !}{6 !}
=[252-56]
\end{array}
\]
\[
=196
\]
Knowledge points:
5.2.2.1 repetition (e.g. the number of ways of arranging the letters of the word ‘NEEDLESS’)
5.2.2.2 restriction (e.g. the number of ways several people can stand in a line if two particular people must, or must not, stand next to each other). (Questions may include cases such as people sitting in two (or more) rows.) (Questions about objects arranged in a circle will not be included.)
Solution:
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