You are given the digits 2, 3, 4, 6, and 7. I | vedclass

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(B) The given digits are $2, 3, 4, 6,$ and $7$. We need to form a $3$-digit number (let the positions be $H, T, U$ for Hundreds,Tens,and Units).For th

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$18$

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(B) The given digits are $2, 3, 4, 6,$ and $7$. We need to form a $3$-digit number (let the positions be $H, T, U$ for Hundreds,Tens,and Units).For the number to be greater than $300$,the hundreds digit $(H)$ must be $3, 4, 6,$ or $7$.For the number to be odd,the units digit $(U)$ must be $3$ or $7$.We analyze based on the choice of $H$:Case $1$: $H = 3$. Then $U$ must be $7$ ($1$ way). The tens digit $(T)$ can be any of the remaining $3$ digits $(2, 4, 6)$. Total $= 1 	imes 3 	imes 1 = 3$.Case $2$: $H = 4$. Then $U$ can be $3$ or $7$ ($2$ ways). The tens digit $(T)$ can be any of the remaining $3$ digits. Total $= 1 	imes 3 	imes 2 = 6$.Case $3$: $H = 6$. Then $U$ can be $3$ or $7$ ($2$ ways). The tens digit $(T)$ can be any of the remaining $3$ digits. Total $= 1 	imes 3 	imes 2 = 6$.Case $4$: $H = 7$. Then $U$ must be $3$ ($1$ way). The tens digit $(T)$ can be any of the remaining $3$ digits $(2, 4, 6)$. Total $= 1 	imes 3 	imes 1 = 3$.Total numbers $= 3 + 6 + 6 + 3 = 18$.

You are given the digits $2, 3, 4, 6,$ and $7$. In how many ways can you form a $3$-digit number without repetition of digits that is odd and greater than $300$?

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