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How many numbers with 6 digits can be formed with the digits $1,2,3,4,5$ such that the digit $2$ appears every time at least three times.

My try:

Total numbers: $5^6$

Numbers in which 2 doesn't appear: $4^6$

Numbers in which 2 appear once : $6\cdot4^5$

Numbers in which 2 appear twice : $13\cdot4^4$

So my result is: $5^6-4^6- 6 \cdot4^5-13\cdot4^4=2057 $ but the right answer is $1545$ How solve it ?

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  • $\begingroup$ Hmm... $2057-1545=2\cdot4^4$ right? $\endgroup$
    – drhab
    Jul 9, 2019 at 9:07
  • $\begingroup$ yes, that means that I lost 2*4^4 numbers $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:10
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    $\begingroup$ Hint: you can insert $2$ elements in $6$ slots in $\binom{6}{2}=15$ different ways $\endgroup$
    – M.P
    Jul 9, 2019 at 9:10
  • $\begingroup$ So evidently you forgot $2$ possibilities. In how many ways can we select $2$ objects out of $6$? $\endgroup$
    – drhab
    Jul 9, 2019 at 9:11
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    $\begingroup$ 6!/(2!*4!)=15 so I forgot 2 cases, thanks $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:12

2 Answers 2

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You are really close to the actual answer here.

Hint: How did you get $13$ in $13\cdot 4^4$?

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  • $\begingroup$ I thought if I have a 6 digit number abcdef the digit 2 must appear in ab ac ad ae af bc bd ....ef (13 cases) so the result is 13 * 4^4 $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:05
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    $\begingroup$ @DaniVaja Count the number of cases more carefully. Note that 5 of them starts with "a", 4 of them starts with "b", and so on. $\endgroup$
    – Arthur
    Jul 9, 2019 at 9:07
  • $\begingroup$ I got 15 cases but using the formula but I don't see those 2 cases in my approach with ab ac ad... $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:15
  • $\begingroup$ Thanks, I didn;t see your edit $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:15
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    $\begingroup$ I totally forgot about de, df.Thank you $\endgroup$
    – DaniVaja
    Jul 9, 2019 at 9:17
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...or the other way around: $\binom{6}{3}4^3 + \binom{6}{4}4^2 + \binom{6}{5}6^1 +1$. Keep in mind $\binom{6}{4} = \binom{6}{2}$. It's a term longer, but gives you an opportunity to compute sm binomial coefficients.

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