The number of odd numbers greater than 6,00,0 | vedclass

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(B) We need to form $6$-digit numbers greater than $6,00,000$ using the digits $\{0, 3, 6, 7, 8, 9\}$ without repetition. For the number to be odd,the

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(B) We need to form $6$-digit numbers greater than $6,00,000$ using the digits $\{0, 3, 6, 7, 8, 9\}$ without repetition. For the number to be odd,the last digit must be $3, 7,$ or $9$ ($3$ choices). For the number to be greater than $6,00,000$,the first digit must be $6, 7, 8,$ or $9$. Case $1$: The first digit is $6$ or $8$ ($2$ choices). The last digit is $3, 7,$ or $9$ ($3$ choices). The remaining $4$ positions can be filled by the remaining $4$ digits in $4! = 24$ ways. Number of ways $= 2 	imes 24 	imes 3 = 144$. Case $2$: The first digit is $7$ or $9$ ($2$ choices). The last digit must be one of the remaining $2$ odd digits ($2$ choices). The remaining $4$ positions can be filled by the remaining $4$ digits in $4! = 24$ ways. Number of ways $= 2 	imes 24 	imes 2 = 96$. Total number of odd numbers $= 144 + 96 = 240$.

The number of odd numbers greater than $6,00,000$ that can be formed by using the digits $3, 6, 7, 8, 9, 0$ without repetition is

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