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I manually calculated that 11 combinations can be made, but I don't know how to do it by maths-formula.

I thought 12 combinations can be made if I apply permutations and combinations as '4C2'= 12.

I found combinations as

1) ab,bc,cd,ad,bd,ac.

2) abc,bcd,abd,acd.

3)abcd.

but these are only 11, Am I missing something or How to calculate it mathematically?

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  • $\begingroup$ Welcome to MSE. Here's some guide to write math in MathJax. $\endgroup$ Jun 20, 2018 at 14:26
  • $\begingroup$ This is not a permutations question unless the order of selection matters. $\endgroup$ Jun 21, 2018 at 9:22

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First note that what you call $4C2$ should come out equal to six, which is the number of combinations of two different letters chosen from four. Various notation is used in different places for the same thing - $^4C_2$ can be used, but most common here (and conveniently available using "\binom 42" with dollar signs instead of quotation marks) is $\binom 42$

As for the answer, each letter can either be included or excluded in the combination, which gives a total of $2^4=16$ possibilities. You have missed out the possibility of no letters at all (which you may not have wanted anyway) and also the four single letters - with the eleven you did find, that gives all sixteen possibilities.

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