The number of 5-digit numbers that can be for | vedclass

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(A) To find the number of $5$-digit numbers formed using digits ${1, 2, 3, 4, 5, 6}$ that include both $1$ and $2$,we use the Principle of Inclusion-E

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$6^5 - 2 \cdot 5^5 + 4^5$

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(A) To find the number of $5$-digit numbers formed using digits ${1, 2, 3, 4, 5, 6}$ that include both $1$ and $2$,we use the Principle of Inclusion-Exclusion.Total $5$-digit numbers using these $6$ digits = $6^5$.Let $A$ be the set of numbers that do not contain $1$,and $B$ be the set of numbers that do not contain $2$.We want to find the total numbers minus those that exclude $1$ or exclude $2$.Number of ways excluding $1$ = $5^5$.Number of ways excluding $2$ = $5^5$.Number of ways excluding both $1$ and $2$ = $4^5$.By the Principle of Inclusion-Exclusion,the number of ways excluding $1$ or $2$ is $|A \cup B| = |A| + |B| - |A \cap B| = 5^5 + 5^5 - 4^5 = 2 \cdot 5^5 - 4^5$.The number of ways including both $1$ and $2$ = Total - $|A \cup B| = 6^5 - (2 \cdot 5^5 - 4^5) = 6^5 - 2 \cdot 5^5 + 4^5$.

The number of $5$-digit numbers that can be formed using the digits $1, 2, 3, 4, 5, 6$ such that the number must include both $1$ and $2$ is:

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