All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Set Me Up (Posted on 2004-02-10)
To demonstrate set union and intersection to her class, Mrs. Putnam asked for three students to each write down a set of numbers.

After they had done so, she looked at their sets and told the class, "the union of these three sets is the first ten counting numbers, but their intersection is empty!"

How many triples (A, B, C) of sets are there such that

A U B U C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
and
A ∩ B ∩ C = {} ?

 See The Solution Submitted by DJ Rating: 4.3636 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution, unless I am overlooking something again :-( | Comment 1 of 31
None of Mrs. Putnam's students would be so geeky as to write down a null set.

Diasallowing null sets, there are 13,680 UNORDERED triples (A, B, C) of sets like these. If the triples are ordered, there are 13,680*(3!) = 82,080 possibilities.

Explanation:

{1 number}U{1 number}U{8 numbers}: 10*9 = 90 unordered possibilities
{1 number}U{2 numbers}U{7 numbers}:
10*(9*8/2!) = 360
{1 number}U{3 numbers}U{6 numbers}:
10*(9*8*7/3!) = 840
{1 number}U{4 numbers}U{5 numbers}:
10*(9*8*7*6/4!) = 1260
{2 numbers}U{3 numbers}U{5 numbers}:
(10*9/2!)*(8*7*6/3!) = 2520
{2 numbers}U{4 numbers}U{4 numbers}:
(10*9/2!)*(8*7*6*5/4!) = 3150
{2 numbers}U{2 numbers}U{6 numbers}:
(10*9/2!)*(8*7/2!) = 1260
{3 numbers}U{3 numbers}U{4 numbers}:
(10*9*8/3!)*(7*6*5/3!) = 4200

90+360+840+1260+2520+3150+1260+4200=13,680

Edited on February 10, 2004, 10:45 am
 Posted by Penny on 2004-02-10 10:20:37

 Search: Search body:
Forums (0)