How to find number of numbers formed with given digits?

Question

Question is Find the number of numbers of five digits that can be made with the digits of the number 1203210. Can you please explain the problem? I did not understand it. Although one asked similar question, I was unable to follow. Can I use multinomial or binary method to solve these problems? Answer is 258

Answer 1

Notice, the significant numbers that can be formed using digits $0$, $1$, $2$ & $3$ should not have first digit $0$. Hence, first place can be filled by any of $\{1, 2, 3\}$ & each of other four places can be filled by 4 ways $\{0, 1, 2, 4\}$

hence assuming that the repetition of digits is allowed, the total 5-digit numbers $$=3\times4\times 4\times 4\times 4=768$$

Answer 2

I think the question means that you can draw digits only from the given #, i.e. two each of 0's, 1's, and 2's and only one of 3.

Remember, a # can't start with a 0.

The solution can easily be obtained using a generating function.

Without restricting 0 at start, numbers possible will be given by

coefficient of $x^5$ in $5!(1+x)(1+x + x^2/2!)^3 = 360$

subtract from this numbers of 4 digits (having only one 0 left), so

coefficient of $x^4 in 4!(1+x)^2(1+x+x^2/2!)^2 = 102

360 - 102 = 258

I can't see an easy way to do it otherwise, but as I said, it is very late here, I'll see later if it is reasonably easy to do it without a G.F.

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The best I can do is to categorise as follows [ revised categorisation ]

(i) Both 0's present and can be placed in ${4\choose 2} = 6 $ ways. [Three from 11223 left to fit in 3 spaces]

Those containing a "double" [112,113,221,223 ] can be permutes in 3 ways each, and 123 can be permuted in 6 ways $\Rightarrow 6\cdot(4\cdot3 +6) = 108$

(ii) One 0 present, placed in 4 ways [Four needed from 11223] 1123 and 2213 can be permuted in 12 ways, while 1122 can be permuted in 6 ways$\Rightarrow 4(2\cdot12 + 6) = 120$

(iii) No 0 present:[5 needed from 11223] $dfrac{5!}{2!2!}=30

Finally, 108+120+30 = 258

Still, I advise use the G.F. Such problems may come in different forms, and it is a tedious error prone process.

Answer 3

A) If the number has no zeros, then there are $\frac{5!}{2!2!}=\color{red}{30}$ possibilities.

B) If the number has exactly 1 zero, then there are $4\binom{4}{2}=\color{red}{24}$ possibilities if it does not contain a 3,

$\hspace{.2 in}$and $2\cdot4\cdot4\cdot3=\color{red}{96}$ possibilities if it does contain a 3

$\hspace{.2 in}$(since there are 2 choices for the repeated digit, and $4\cdot4\cdot3$ ways to arrange the digits).

C) If the number has 2 zeros, then there are $\binom{4}{2}\cdot6=\color{red}{36}$ possibilities if 0 is the only repeated digit

$\hspace{.2 in}$and $2\cdot2\cdot\binom{4}{2}\cdot3=\color{red}{72}$ possibilities if there is another repeated digit

(since there are 2 choices for the other repeated digit, 2 choices for the 3rd digit, $\binom{4}{2}$ ways to place the zeros, and 3 ways to place the nonrepeated digit).

This gives a total of $\color{red}{258}$ possibilities.

Answer 4

A way of computing the number of "words" you can make out of MISSISSIPPI ($4 \times I, 1 \times M, 4 \times S, 2 \times P$) is to take the $4 + 1 + 4 + 2 = 11$ letters with subindices to distinguish equal letters, i.e., consider permuting $I_1 I_2 I_3 I_4 M S_1 S_2 S_3 S_4 P_1 P_2$, which can be done in $11!$ ways. But if you erase the indices from the $I$s, you see that a factor of $4!$ alternative words dissapear. Do the same for all, you end up with:

$$ \frac{11!}{4!\,1!\,4!\,2!} = \binom{11}{4, 1, 4, 2} $$

a multinomial coefficient.

In your case, there are $2 \times 0, 2 \times 1, 2 \times 2, 1 \times 3$, this would give:

$$ \binom{2+2+2+1}{2, 2, 2, 1} = 630 $$

But presumably starting with $0$ is forbidden. One way to handle that is to consider you have to distribute the two zeros in the 6 positions that aren't the first one (for

$$ \binom{6}{2} = 15 $$

options), and compute the ways of shuffling the rest of the digits in order to the still free positions (there are

$$ \binom{2+2+1}{2,2,1} = 30 $$

ways to do this). The decision of placing the $0$ and the other digits are independent, so the total is:

$$ \binom{6}{2} \cdot \binom{2+2+1}{2,2,1} = 15 \cdot 30 = 450 $$