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

now consider


now first place cannot be filled with 0. so number of ways of filling first blank is $\frac{6!}{2! 3!}$. how do i go furthure?

thanks

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  • $\begingroup$ You mean how many 7 digit numbers? $\endgroup$ – marshal craft Apr 10 '17 at 10:49
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Count all 7 digit numbers that can be formed using the 7 digits, and then subtract the count of those starting with zero: $$\frac{7!}{2!3!}-\frac{6!}{2!3!} $$

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  • $\begingroup$ you have included numbers starting with 0 in 7 digit numbers ? how? $\endgroup$ – J. Deff Apr 10 '17 at 10:53
  • $\begingroup$ Represent the digits say the letters of their French names z, d,d,t,t,t,q: So you want to count how many words of length $7$ can be formed using them but not starting with z. $\endgroup$ – P Vanchinathan Apr 10 '17 at 10:59
  • $\begingroup$ its quite clear. is there other way to do this? $\endgroup$ – J. Deff Apr 10 '17 at 11:06
  • $\begingroup$ I tried with German. But zwei starting with z confused me with zero. :-) $\endgroup$ – P Vanchinathan Apr 10 '17 at 11:08

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