How many numbers greater than a million can b | vedclass

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(B) The given digits are $2, 3, 0, 3, 4, 2, 3$. There are $7$ digits in total,where $2$ appears twice,$3$ appears thrice,$0$ appears once,and $4$ appe

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$360$

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(B) The given digits are $2, 3, 0, 3, 4, 2, 3$. There are $7$ digits in total,where $2$ appears twice,$3$ appears thrice,$0$ appears once,and $4$ appears once.All $7$-digit numbers formed using these digits are greater than a million.The digit at the millions place cannot be $0$.Therefore,the millions place can be filled by $2, 3,$ or $4$.Case $I$: Millions place is $2$.Remaining digits are $3, 0, 3, 4, 2, 3$. The number of arrangements is $\frac{6!}{3!} = \frac{720}{6} = 120$.Case $II$: Millions place is $3$.Remaining digits are $2, 0, 3, 4, 2, 3$. The number of arrangements is $\frac{6!}{2! 	imes 2!} = \frac{720}{4} = 180$.Case $III$: Millions place is $4$.Remaining digits are $2, 3, 0, 3, 2, 3$. The number of arrangements is $\frac{6!}{3! 	imes 2!} = \frac{720}{6 	imes 2} = 60$.Total numbers $= 120 + 180 + 60 = 360$.

How many numbers greater than a million can be formed with the digits $2, 3, 0, 3, 4, 2, 3$?

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