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

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Question

(A) Let $S$ be the set of all $5$-digit numbers formed using digits ${1, 2, 3, 4, 5, 6}$. The total number of such numbers is $6^5$.Let $A$ be the set

Vedclass · Accepted Answer

$6^5 - 2 	imes 5^5 + 4^5$

Vedclass · Answer

(A) Let $S$ be the set of all $5$-digit numbers formed using digits ${1, 2, 3, 4, 5, 6}$. The total number of such numbers is $6^5$.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 number of $5$-digit numbers that contain both $1$ and $2$. This is given by: Total - (Numbers without $1$ $OR$ numbers without $2$).Using the Principle of Inclusion-Exclusion,$|A \cup B| = |A| + |B| - |A \cap B|$.$|A| = 5^5$ (numbers formed using ${2, 3, 4, 5, 6}$).$|B| = 5^5$ (numbers formed using ${1, 3, 4, 5, 6}$).$|A \cap B| = 4^5$ (numbers formed using ${3, 4, 5, 6}$).Thus,$|A \cup B| = 5^5 + 5^5 - 4^5 = 2 	imes 5^5 - 4^5$.The number of $5$-digit numbers containing both $1$ and $2$ is $6^5 - (2 	imes 5^5 - 4^5) = 6^5 - 2 	imes 5^5 + 4^5$.

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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