Three and four digit numbers are formed from | vedclass

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(C) Given digits are $\{1, 3, 5, 6, 8\}$. Even digits are $\{6, 8\}$ and odd digits are $\{1, 3, 5\}$.$\bullet$ Calculation of $e_1$ ($3$-digit even n

Vedclass · Accepted Answer

$\frac{e_1+e_2}{2}=\frac{O_1+O_2}{3}=6^2$

Vedclass · Answer

(C) Given digits are $\{1, 3, 5, 6, 8\}$. Even digits are $\{6, 8\}$ and odd digits are $\{1, 3, 5\}$.$\bullet$ Calculation of $e_1$ ($3$-digit even numbers,no repetition):The last digit must be even ($2$ choices: $6$ or $8$). The remaining $2$ places can be filled by the remaining $4$ digits in $P(4, 2)$ ways.$e_1 = 2 	imes (4 	imes 3) = 24$.$\bullet$ Calculation of $e_2$ ($4$-digit even numbers,no repetition):The last digit must be even ($2$ choices). The remaining $3$ places can be filled by the remaining $4$ digits in $P(4, 3)$ ways.$e_2 = 2 	imes (4 	imes 3 	imes 2) = 48$.$\bullet$ Calculation of $O_1$ ($3$-digit odd numbers,no repetition):The last digit must be odd ($3$ choices: $1, 3, 5$). The remaining $2$ places can be filled by the remaining $4$ digits in $P(4, 2)$ ways.$O_1 = 3 	imes (4 	imes 3) = 36$.$\bullet$ Calculation of $O_2$ ($4$-digit odd numbers,no repetition):The last digit must be odd ($3$ choices). The remaining $3$ places can be filled by the remaining $4$ digits in $P(4, 3)$ ways.$O_2 = 3 	imes (4 	imes 3 	imes 2) = 72$.Now,check the options:$\frac{e_1+e_2}{2} = \frac{24+48}{2} = \frac{72}{2} = 36$.$\frac{O_1+O_2}{3} = \frac{36+72}{3} = \frac{108}{3} = 36$.Since $36 = 6^2$,the correct relation is $\frac{e_1+e_2}{2} = \frac{O_1+O_2}{3} = 6^2$.

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