# Distinct numbers formed by using six digits

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?

• $P^3_2 = 3 \times 2 = 6$, 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. – antkam Mar 14 at 18:12
• 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. – M. Kumar Mar 14 at 18:28
• Yep, I think your last comment got it right. – antkam Mar 14 at 18:32
• 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. – user Mar 14 at 20:45
• Are you sure? It's not given in the question that how many digits the formed number should have. – M. Kumar Mar 15 at 6:48

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.

• 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. – M. Kumar Mar 15 at 6:44
• 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). – Wolfgang Kais Mar 15 at 8:51