Given:
Number of red flags \[=\text{ }4\]
Number of white flags \[=\text{ }2\]
Number of green flags \[=\text{ }3\]
So there are total \[9\] flags, out of which \[4\] are red, \[2\] are white, \[3\]are green
By using the formula,
\[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\text{ }=\text{ }9!\text{ }/\text{ }\left( 4!\text{ }2!\text{ }3! \right)\]
\[=\text{ }\left[ 9\times 8\times 7\times 6\times 5\times 4! \right]\text{ }/\text{ }\left( 4!\times 2\times 1\times 3\times 2\times 1 \right)\]
Or,
\[=\text{ }\left[ 9\times 8\times 7\times 6\times 5 \right]\text{ }/\text{ }\left( 2\times 3\times 2 \right)\]
\[=\text{ }9\times 4\times 7\times 5\]
So,
\[=\text{ }1260\]
Hence, \[1260\] different signals can be made.