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Suppose I have five numbers: 12345. I want to split them into 2, 3, 4 and so on possible ways:

If we split into two columns, we have 4 these possibilities:

1    | 2345
12   | 345
123  | 45
1234 |  5

If we split into three columns, we have 6 these possibilities:

 1 | 2   | 345
 1 | 23  | 45
 1 | 234 |  5
12 | 3   | 45
12 | 34  | 5
123| 4   | 5

If split into four columns, we have 4 these possibilities:

12 |  3 |  4 | 5
 1 | 23 |  4 | 5
 1 |  2 | 34 | 5
 1 |  2 |  3 | 45

Split into five columns, we have 1 possibility

1 | 2 | 3 | 4 | 5

So, we can split 5 numbers into maximum 5 columns and In total we have 15 possibilities we can split these up.

Just as this representation shown above, is there a mathematical way to count for these possibilities for any number?

cpx
  • 639

1 Answers1

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Hint: If you include the "split" into one column, you have a total of $16=2^{5-1}$ possibilities.

Barry Cipra
  • 79,832
  • What about if I want to know e.g. in how many ways I can split 12345 into 3 columns as above? Is there a certain rule or concept of maths we are following or we are making it up our own? – cpx Apr 23 '14 at 14:49
  • @cpx, to split an ordered list of $n$ objects into $k$ columns, you need to insert $k-1$ column markers in the $n-1$ gaps between consecutive objects. Note that ${5-1\choose3-1}=6$. – Barry Cipra Apr 23 '14 at 14:55
  • I couldn't have figured. Just to make sure, are those binomial coefficient in big bracket? – cpx Apr 23 '14 at 15:12
  • Yes, those are binomial coefficients. – MJD Apr 23 '14 at 15:43
  • @cpx, yes, the key idea is that you get to choose where to put the column markers. (A binomial coefficient $n\choose k$ is normally read as "$n$ choose $k$.") – Barry Cipra Apr 23 '14 at 15:55