Number of all possible words (with or without | vedclass

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(B) The word $CABINET$ has $7$ distinct letters: $C, A, B, I, N, E, T$. The total number of arrangements is $7! = 5040$. Let $S$ be the set of all arr

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$4806$

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(B) The word $CABINET$ has $7$ distinct letters: $C, A, B, I, N, E, T$. The total number of arrangements is $7! = 5040$. Let $S$ be the set of all arrangements. $|S| = 5040$. Let $X$ be the set of arrangements containing $CAB$. Treating $CAB$ as one unit,we have $5$ units: $(CAB), I, N, E, T$. These can be arranged in $5! = 120$ ways. Let $Y$ be the set of arrangements containing $NET$. Treating $NET$ as one unit,we have $5$ units: $C, A, B, I, (NET)$. These can be arranged in $5! = 120$ ways. Let $X \cap Y$ be the set of arrangements containing both $CAB$ and $NET$. Treating $CAB$ and $NET$ as two units,we have $2$ units: $(CAB), I, (NET)$. These can be arranged in $3! = 6$ ways. By the Principle of Inclusion-Exclusion,the number of arrangements containing $CAB$ or $NET$ is $|X \cup Y| = |X| + |Y| - |X \cap Y| = 120 + 120 - 6 = 234$. The number of arrangements in which neither $CAB$ nor $NET$ appear is $|S| - |X \cup Y| = 5040 - 234 = 4806$.

Number of all possible words (with or without meaning) that can be formed using all the letters of the word $CABINET$ in which neither the word $CAB$ nor the word $NET$ appear is

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