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I'm looking for $8$ digit numbers with $4$ digits being used twice. for example : $11223344$ and $12123434$ and $11002233$

Its not allowed to use one digit like: $11223345$

for $4$ digit numbers with $2$ digits being used twice I have computed $243$ numbers.

I need to find out how many 8-digit numbers satisfy afore-mentioned conditon

MasOOd.KamYab
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4 Answers4

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The number of ways to pair up the digit positions is $7 \cdot 5 \cdot 3 \cdot 1 = 105$. For each such pairing, to get a number you assign the first pair (including the leading digit) any of the $9$ possible digits $1,2,\ldots,9$ (since we don't want a leading $0$), the next pair any of $9$ remaining possibilities (including $0$ this time, but not the one assigned to the first pair), the next any of $8$, and the last pair any of $7$. Thus there are $105 \cdot 9 \cdot 9 \cdot 8 \cdot 7 = 476280$ possible numbers.

Robert Israel
  • 448,999
  • can you explain for 4-digit numbers with 2 digits being used twice? – MasOOd.KamYab May 13 '20 at 19:44
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    $3$ pairings so $3 \times 9 \times 9 = 243$. – Robert Israel May 13 '20 at 19:59
  • Sir, could you explain more about the multiplication you used for pairing digit position? – MasOOd.KamYab May 13 '20 at 20:16
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    The first position can be paired with any of the remaining $7$. Remove these two and you have $6$ still unpaired. The next yet-unpaired position is paired with any of the remaining $5$. That leaves $4$. The next yet-unpaired position is paired with any of the remaining $3$. That leaves $2$, which must be paired with each other. – Robert Israel May 14 '20 at 02:05
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The first digit on the left can not be $0$. So there are $9$ possible digits the first digit can be. Call that first digit $a$.

You will need another $a$ somewhere. There are $7$ places to put the second $a$.

Now you need a digit to go in the first free space from the left. The digit can be anything except $a$. Ao there are $9$ possible digits for the digit in the first free space to be. Call that digit $b$.

You will need another $b$ somewhere. There are $5$ places to put the second $b$.

Now you need a digit to go in the first free space from the left. There are $8$ possible digits. Call it $c$.

Now you need to place a second $c$ somewhere. There are $3$ places left.

Now at this point there are only two spaces left. They must both have the same digit. There are $7$ choices for that last digit.

So there are $9*7*9*5*8*3*7$ possible numbers.

I was very strict on order because I wanted to avoid double counting or any potential taking casese into accont.

fleablood
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If $0$ is not included, then we have $9\choose 4$ ways to choose the pairs, and $\frac{8!}{(2!)^4}$ ways to arrange them. If $0$ is included, then there are $9\choose 3$ choices for the other three pairs. Now, the two $0's$ can be arranged in $7\choose 2$ ways, and for the remaining digits we have $\frac{6!}{(2!)^3}$ ways of permuting them. This gives the total ways: $${9\choose 4} \times \frac{8!}{(2!)^4} + {9\choose 3}\times {7\choose 2}\times \frac{6!}{(2!)^3}=\boxed{476280}$$

Vishu
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Case 1: $0$ is not included.

Out of nine numbers i.e (1,2,..., 9), selecting any four numbers by $$9\choose4$$ ways and rearranging them in $$\frac{8!}{(2!)^4}$$

$\therefore$ Total numbers=$${9\choose4}{\frac{8!}{(2!)^4}}=317520$$

Case 2: $0$ is included.

Now, two places are reserved for $0$. So we want three numbers to fill six places. That can be done in $9\choose3$ ways. Also $0$ can't occur at first place ($\because$ it will be then $7$ digit number).

They can be arranged in $$\frac{8!}{(2!)^4}-\frac{7!}{(2!)^3}$$ ways.

$\therefore$ Total numbers=$${9\choose3}\left(\frac{8!}{(2!)^4}-\frac{7!}{(2!)^3}\right)=158760$$

Now add the numbers of both the cases to get the final answer as $476280$.

SarGe
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