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How many different numbers can be formed by various arrangements of the six digits 1, 1, 1, 1, 2, 3?

My attempt: Numbers formed by the given digits can have 1 to 6 digits. There are 3, 7, $^3P_2$, $^4P_2$, $^5P_2$ & $^6P_2$ possibilities for 1,2,3,4,5 and 6 digit numbers respectively.
So, the answer is the sum: $3+ 7+ ^3P_2+ ^4P_2+ ^5P_2+^6P_2$

Is this correct?

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  • $\begingroup$ $P^3_2 = 3 \times 2 = 6$, or am I misinterpreting you? Anyway there are more than $6$ 3-digit numbers that can be formed according to your rule. There are $6$ just from permuting $123$, and then there are $3$ from permuting $112$, $3$ from permuting $113$, and $1$ more from $111$, for a total of $6+3+3+1 = 13$ if I'm not missing anything. $\endgroup$ – antkam Mar 14 at 18:12
  • $\begingroup$ Yeah, you are right. I think it would be $3$, $^2P_2+2+2+1$, $^3P_2+3+3+1$, $^4P_2+4+4+1$, $^5P_2+5+5$ & $^6P_2$ possibilities for 1,2,3,4,5 and 6 digit numbers respectively. $\endgroup$ – M. Kumar Mar 14 at 18:28
  • $\begingroup$ Yep, I think your last comment got it right. $\endgroup$ – antkam Mar 14 at 18:32
  • $\begingroup$ If you write about an arrangement of six digits, you cannot count numbers having less digits. It is especially obvious if a question lists a digit 4 times. $\endgroup$ – user Mar 14 at 20:45
  • $\begingroup$ Are you sure? It's not given in the question that how many digits the formed number should have. $\endgroup$ – M. Kumar Mar 15 at 6:48
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You could partition the problem into smaller problems composing numbers from the 6 digits:

  1. How many different numbers contain neither $2$ nor $3$
  2. How many different numbers contain just one of $2$ and $3$
  3. How many different numbers contain both $2$ and $3$

Each of these numbers can be obtained by counting the solutions by the number of $1$'s they contain:

  1. The composed string of digits will consist of one up to four $1$'s, there are $4$ ways to do so.
  2. There are $2$ ways to decide between $2$ and $3$, and for $k=0,...,4$ there are $k+1$ ways to place that digit between $k$ $1$'s. So, counting the possibilities, we get $2\sum_{k=0}^{4}(k+1) = 2\sum_{k=0}^{4}\binom{k+1}{1} = 2\binom{4+2}{2} = 30$.
  3. For $k=0,...,4$, starting with $k$ $1$'s, we have $k+1$ possible positions to insert the $2$ and then $k+2$ possible positions to place the $3$. Summing them up, we get $\sum_{k=0}^{4}(k+1)(k+2) = \sum_{k=0}^{4}\binom{k+2}{2}2! = 2\binom{4+3}{3} = 70$.

In total, we get $4+30+70 = 104$ possible ways to compose the number.

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  • $\begingroup$ I think this is the same thing as I have done in my first comment on the question, but in a well arranged manner. Please check it once. $\endgroup$ – M. Kumar Mar 15 at 6:44
  • $\begingroup$ Well, your formula contains $6$ summands (the number of digits) while mine contains just $3$ (the number of distinct digits), corrected by $1$. You might have not seen the well-arrangedness of my solution. In fact it is $$ \sum_{k=0}^{n}k!\binom{n}{k}\binom{m+1+k}{m} - 1 $$ where $n$ is the number of "additional digits" ($2$ in your case) and $m$ is the number of $1$'s ($4$ in your case). $\endgroup$ – Wolfgang Kais Mar 15 at 8:51

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