I have the following question. I have a set of elements $A,B,C$ like this: $$ A_nB_mC_l $$ $n,m,l$ are the element counts.

How to calculate all possible combinations (without replacement, the order is not important) according to $n,m,l$?

For example if I have $A_2B_2C$ there are following combinations possible:

amount of elements = 1: $A,B,C$ - 3 combinations

amount of elements = 2: $A_2,B_2,AB,AC,BC$ - 5 combinations

amount of elements = 3: $A_2B,A_2C,AB_2,B_2C,ABC$ - 5 combinations

amount of elements = 4: $A_2BC,AB_2C,A_2B_2$ - 3 combinations

amount of elements = 5: $A_2B_2C$ - 1 combination

So, in total I have $3+5+5+3+1=17$ combinations

  • $\begingroup$ (n+1)*(m+1)*(l+1)-1 ................... $\endgroup$
    – hvedrung
    Commented Oct 9, 2014 at 13:05
  • $\begingroup$ You have omitted one possibility: amount of elements = 0: one combination (the empty set). Including this will help you derive a simpler formula, because all you have to do to identify a "combination" is to stipulate the number of $A$'s (between $0$ and $n$ inclusive), the number of $B$'s (between $0$ and $m$ inclusive), and the number of $C$'s (between $0$ and $l$ inclusive). $\endgroup$
    – whuber
    Commented Oct 9, 2014 at 18:38

2 Answers 2


You can see it this way

Every tuplet (a, b ,c) such as $a\in[0,n], b\in[0,m], c\in[0,l] $ maps to exactly one combination, and every combination maps to exactly one tuplet

that means that: $F(a,b,c) => A_aB_bC_c$ is a bijection between [0,n][0,m][0,l] and the set of all possible combinations hence both sets have the same cardinality.

The cardinatlity of [0,n][0,m][0,l], and hence of the number of possible combinations is (n+1)(m+1)(l+1)


I guess this can be seen as a simple example: Consider A and B and C as cities and m and n and l as different roads between these three cities respectively. the question can be pictured as following:

How many ways that a traveler can travel between these 3 cities, assuming that he or she can travel directly from A to C.

this can be answered as first, how many combinations of traveling choices?

ABC, AB, AC, BC and staying at either of A, B or C(just going through the paths and coming back to the city).

now, for traveling through ABC, we have k=m X n X l for AB we have q=m X n choices which is also equal to 4 for AC we have w=m X l choices which is equal to 2 for BC we have p=n X l choices which is equal to 2 and finally for staying in either cities , we have m and n and l choices.

the final answer is : k+q+w+p+m+n+l


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.