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How many seven - digit even numbers greater than $4,000,000$ can be formed using the digits $0,2,3,3,4,4,5$?

I have solved the question using different cases: when $4$ is at the first place and when $5$ is at the first place, then using constraints on last digit.

But is there a smarter way ?

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Sorry I was misreading your efforts. I think your way is straight forward. BTW: Is this homework, if so please tag it accordingly? –  draks ... Jul 23 '12 at 10:11
    
thank you for this –  Arpit Bajpai Jul 23 '12 at 14:22

3 Answers 3

up vote 1 down vote accepted

I think considering the two different cases you mentioned separately is best. To avoid further case division, I'd proceed like this:

  • Case with leading digit 4: since an even digit has to go in the rightmost position, there are $5 \choose 3$ ways to choose the positions of the 3 odd numbers amongst the other 5 positions, then 3 ways to order them, and $3!$ ways to order the even numbers, for 180 total.

  • Case with leading digit 5: similarly, there are $5 \choose 2$ ways to choose the positions of the two odd digits (both 3's), only 1 way to order them, then $\frac{4!}{2}$ ways to order the even digits (dividing by 2 since there are 2 indistinguishable 4's), for 120 total.

So all up there are 300 such numbers.

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Combinations can be:

Case1#: $5 \{4 3 3 2 4\} 0$, total combinations are: $\frac{5!}{2!2!} = 30$

Case2#: $4 \{5 3 3 2 4\} 0$, total combinations are: $\frac{5!}{2!} = 60$

Case3#: $5 \{4 3 3 0 4\} 2$, total combinations are: $\frac{5!}{2!2!} = 30$

Case4#: $4 \{5 3 3 0 4\} 2$, total combinations are: $\frac{5!}{2!} = 60$

Case5#: $5 \{4 3 3 2 0\} 4$, total combinations are: $\frac{5!}{2!} = 60$

Case6#: $4 \{5 3 3 2 0\} 4$, total combinations are: $\frac{5!}{2!} = 60$

So, the total no of combination will be $300$.

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You have the following combinations:

  • $4\dots 0$, where in the middle you permute $\{2,3,3,4,5\}: \frac{5!}{2}$, with the $2$ in the denominator accounts for the double $3$. And the same for $4\dots 2$, $4\dots 4$ and $5\dots 4$.
  • $5\dots 0$, where in the middle you permute $\{2,3,3,4,4\}: \frac{5!}{2^2}$, with 2 doubles $3$ and $4$. And the same for $5\dots 2$.

So in total you have $4\frac{5!}{2}+2\frac{5!}{2^2}=300\;$ even numbers.

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