number of permutation of 9 digit numbers (1 Viewer)

juantheron · Mar 1, 2016

$Number of permutations of $1,2,3,4,5,6,,7,8,9$ such that $1$ appear$

$somewhere to the left of $2$ and $3$ appear to the left of $4$ and $5$

$somewhere to the left of $6,$ are$

$like $815723946$ would be one such permutation$

Paradoxica · Mar 1, 2016

We can ignore 7,8,9 without loss of generality, as they do not exert any restraints on the ordered pairs. They contribute a factor of 7 X 8 X 9 possibilities to the total number of sequences possible. Without loss of generality, fix the order of the digit pairs.

The first position must contain an odd number. Let this odd number be A, and it's even counterpart be A*. Then there are five possible cases for the position of A*. In each case, the next odd number, B, will follow in either the second or third position. The third position can only occur in the case where A is immediately followed by A* . Then in each case, B* has only three possible positions. With the remaining two slots, there is only one way C and C* can be placed to complete the six-digit number. Thus, the total number of possibilities is 9 X 8 X 7 X 5 X 3 X 3! = 45360

InteGrand · Mar 1, 2016

juantheron said:
$Number of permutations of $1,2,3,4,5,6,,7,8,9$ such that $1$ appear$

$somewhere to the left of $2$ and $3$ appear to the left of $4$ and $5$

$somewhere to the left of $6,$ are$

$like $815723946$ would be one such permutation$

Here is another way of going about it.

$\noindent The answer is $\binom{9}{2}\cdot \binom{7}{2} \cdot \binom{5}{2}\cdot 3!$ (which evaluates to $45360$ as Paradoxica had). The reasoning is that we first choose spots for $1$ and $2$ (note that once we choose the spots, we have automatically chosen the left one of these to contain the 1, so this makes 1 be left of 2). Then from the remaining seven spots, we choose where to put the 3 and 4, done in $\binom{7}{2}$ ways (again, we note that once the spots are chosen, we automatically place 3 in the left one). Then we place 5 and 6 in their spots, and their spots are chosen from the remaining five spots in $\binom{5}{2}$ ways. After this, there are three numbers left, which can be placed in any of the possible $3!$ orders for them, since we don't have any restrictions on them. So the answer is $\binom{9}{2}\cdot \binom{7}{2} \cdot \binom{5}{2}\cdot 3!=45360$.$

number of permutation of 9 digit numbers (1 Viewer)

juantheron

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Paradoxica

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InteGrand

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