MaplePrimes - answers and comments on Question, Combinatorial problem

Solution

Kitonum — Tue, 05 Mar 2013 20:44:44 Z

P:=combinat[powerset]({a, b, c, d, e}):

m:=0: n:=0: k:=0:

for i in P do

for j in P do

if nops(i union j)=5 and nops(i intersect j)=3 then m:=m+1: fi:

if nops(i union j)=5 and nops(i intersect j)=4 then n:=n+1: fi:

if nops(i union j)=5 and nops(i intersect j)=5 then k:=k+1: fi:

od: od:

binomial(30, 5)*(m+n*binomial(4, 3)+k*binomial(5, 3))*2^(30-3);

1721414839173120

Manually

Joe Riel — Tue, 05 Mar 2013 21:38:03 Z

Computing manually I get 30!/25!/3!/2!*5^2 = 35626500. This comes from creating the venn diagram of the sets, noting that there must be exactly 3 in the middle region, 25 in C minus A minus B, leaving 2 for the other 5 regions. Those two can be distributed 5^2 ways. The lhs of the product is the number of ways to partition a set of 30 elements in subsets of 25, 3, and 2.

Note: fixed original computation

Per response from OP, my assumption that A union B union C = S is incorrect. Fortunately, it is easy to handle the more general case, just multiply by the number of ways to partition a set of 25 elements into two sets (the sets being C \ (A union B) and S \ (A union B union C)). That factor is 2^25. So the overall result is

30!/25!/3!/2!*5^2*2^25 = 1195426971648000

Split cases via |A intersect B| = 3, 4, or 5.

Carl Love — Wed, 06 Mar 2013 07:41:28 Z

Notation: I'll use juxtaposition to indicate intersection, `+` for union, `-` for minus, with `+` having precedence. |A| is nops(A), and (n C k) is a binomial coefficient.

We split cases via |AB|. There are three cases, |AB| = 3, 4, or 5. Regardless of the case, we have a factor of (30 C 5) for the choice of A+B and a factor of 2^25 choices for C - A+B.

Case |AB| = 5: Thus A=B. There are (5 C 3) = 10 choices for ABC in AB.

Case |AB| = 4: Either |A|=4 and |B|=5 or vice versa. This dichotomy contributes a factor of 2, and we will consider only the |A|=4 case without loss of generality. There are (5 C 4) choices of A within A+B and (4 C 3) choices of ABC in A. Total factor for this case 2*(5 C 4)*(4 C 3) = 40.

Case |AB| = 3:

Subcase |A|=3, |B|=5 or vice versa. Factor of 2 for the dichotomy. (5 C 3) choices of A in A+B, and only 1 choice of ABC in A. Total factor: 20
Subcase |A|=4, |B|=4. (5 C 4) choices of A in A+B, (4 C 3) choices of AB in A, and(3 C 3) choices of ABC in AB, for a total of 20.

Grand total: (20+20+40+10)*(30 C 5)*2^25 = 430353709793280 =

Edit: Corrections to the above.

Case |AB| = 3:

Subcase |A|=3, |B|=5 or vice versa. Factor of 2 for the dichotomy. (5 C 3) choices of A in A+B, and only 1 choice of ABC in A. For each of the 2 elements of B-A, there is a choice of whether it is in C, for a factor of 2^2. Total factor: 2*(5 C 3)*2^2 = 80.
Subcase |A|=4, |B|=4. (5 C 4) choices of A in A+B, (4 C 3) choices of AB in A, and(3 C 3) choices of ABC in AB. For each of the 2 elements of (A-B)+(B-A), there is the choice of whether it is in C. Total factor: (5 C 4)*(4 C 3)*2^2 = 80.

Grand total: (10+80+80+80)*(30 C 5)*2^25 = =

Want for explanation

Markiyan Hirnyk — Tue, 05 Mar 2013 20:57:59 Z

Can you kindly explain your code? Especially, why is every C counted only one time? I also think that the condition nops(i) >= nops (j) is necessary to get rid of duplications.

Explanation

Kitonum — Tue, 05 Mar 2013 21:44:58 Z

@Markiyan Hirnyk

Let A union B ={a, b, c, d, e} is a subset of S. Then A and B are subsets of {a, b, c, d, e} . From the second condition about the intersection follows that nops(A intersect B)>=3 . Suppose nops(A intersect B)=3 . The number of such pairs A and B is m (see code). Set C, which intersects with (A intersect B) in these 3 elements can be chosen in 2^(30-3) ways and so on.

Possible duplications

Markiyan Hirnyk — Tue, 05 Mar 2013 21:53:29 Z

@Kitonum Why [A, B, C] is counted only one time? It is not obvious if nops (A intersect B) = 4 or 5.