(i) \[INDEPENDENCE\]
There are \[12\] letters in the word \[INDEPENDENCE\] out of which \[2\text{ }are\text{ }Ds,\text{ }3\text{ }are\text{ }Ns,\text{ }4\text{ }are\text{ }Es\]and the rest all are distinct.
So by using the formula,
\[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\]
total number of arrangements \[=\text{ }12!\text{ }/\text{ }\left( 2!\text{ }3!\text{ }4! \right)\]
\[=\left[ 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right]/\left( 2!3!4! \right)\]
\[=\text{ }\left[ 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5 \right]/\left( 2\times 1\times 3\times 2\times 1 \right)\]
Or,
\[=\text{ }11\times 10\times 9\times 8\times 7\times 6\times 5\]
\[=\text{ }1663200\]
(ii) \[INTERMEDIATE\]
There are \[12\] letters in the word \[INTERMEDIATE\] out of which \[2\text{ }are\text{ }Is,\text{ }2\text{ }are\text{ }Ts,\text{ }3\text{ }are\text{ }Es\] and the rest all are distinct.
So by using the formula,
\[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\]
total number of arrangements \[=\text{ }12!\text{ }/\text{ }\left( 2!\text{ }2!\text{ }3! \right)\]
\[=\text{ }\left[ 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right]\text{ }/\text{ }\left( 2!\text{ }2!\text{ }3! \right)\]
Or,
\[=\text{ }\left[ 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 3\times 2\times 1 \right]\text{ }/\text{ }\left( 3! \right)\]
\[=\text{ }12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\]
So,
\[=\text{ }19958400\]