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Find the number of five digit numbers that can be formed by using two 2's,three 3's,one zero,one 5.


Five digit numbers are 33350 and its arrangements,
33322 and its arrangements,
22503 and its arrangements,
33502 and its arrangements,
33302 and its arrangements,
33352 and its arrangements,
33220 and its arrangements.

So the total numbers are $\frac{5!}{3!}+\frac{5!}{3!2!}+\frac{5!}{2!}+\frac{5!}{2!}+\frac{5!}{3!}+\frac{5!}{3!}+\frac{5!}{2!2!}=220$ but the correct answer is $212.$

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    $\begingroup$ You seem to be ignoring the fact that $0$ cannot be the first digit in a $5$ digit number. (don't confuse a 5-digit sequence or string with a 5-digit number). $\endgroup$ – JMoravitz Jul 20 '17 at 1:56
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    $\begingroup$ Are you forgetting that five digit numbers can't begin with $0$? $\endgroup$ – lulu Jul 20 '17 at 1:56
  • $\begingroup$ are you thinking, about how at least one number has to be left out, as the selections you allow number 6. $\endgroup$ – user451844 Jul 20 '17 at 2:08
  • $\begingroup$ I got the answer,i realised my mistake.Thanks. $\endgroup$ – learner_avid Jul 20 '17 at 2:13

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