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All possible combinations of 4 numbers from 0 9 with repetition

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Answered by kinjal9004840142
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kinjal9004840142

Secondary SchoolMath 13+7 pts



All possible combinations of 4 numbers from 0 9 with repetition

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kinjal9004840142

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stylish5551

Stylish5551 Ace

Yes, there does exist such a way. First you select a digit d from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then you select a digit e from ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d). Then you select a digit f from (({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d)-e). You first select 0 for d, then 1, and so on until you get to 7. And, you always select the least digit first for e and f also, with the additional condition that d < e < f. List out the first sequence, 012, 013, 014, 015, 016, 017, 018, 019. Then list all the other numbers beneath them with the condition that for all numbers e and f, and with d held constant, the digits for e and f follow the natural number sequence down the column. Partition each set of sequences by d. The column rule only applies within each partition. (this description might come as incomplete or could use some revision).


The list thus goes:


012, 013, ..., 019 023, 024, ..., 029 034, 035, ..., 039 . . . 089 123, 124, ..., 129 134, 135, ..., 139 . . . 189 . . . 789


I'll clarify the last part here:


567, 568, 569 578, 579 589 678, 679 689 789


Perhaps better, say we try to do the same thing in base 4. The entire sequence goes


012, 013 023 123


In base 5 the process yields:


012, 013, 014 023, 024 034 123, 124 134 234


Thus, in base 10 the sum of the first 8 triangular numbers gives us the number of such combinations: +(1, 3, 6, 10, 15, 21, 28, 36)=120. Similarly, it should logically follow that for x digit numbers in base z, where x < z, or x=z, there exist +[T11, ..., T(z−(x+1))(z−(x+1))] such combinations, where Tnn indicates the nth triangular number.


I guess the underlying idea I've used here lies in following the natural number sequence across the rows, and down the columns for the digits e and f also.

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