2006-11-01 07:04:46 · 4 answers · asked by Sonja - 1 in Science & Mathematics ➔ Mathematics
If you include combinations with no people and all people, the answer is 32 ways. 2^5.
{ } = 1
{A}, {B}, {C}, {D}, {E} = 5
{AB}, {AC}, {AD}, {AE}, {BC}, {BD}, {BE}, {CD}, {CE}, {DE} = 10
{ABC}, {ABD}, {ABE}, {ACD}, {ACE}, {ADE}, {BCD}, {BCE}, {BDE}, {CDE} = 10
{ABCD}, {ABCE}, {ABDE}, {ACDE}, {BCDE} = 5
{ABCDE} = 1
That's a total of 32 ways.
Another way to look at is to think that each person can be included or excluded. There are two ways to include/exclude person A, two ways to include/exclude person B, etc. That's 2^5 = 32.
This all assumes that ordering doesn't matter (that a combination of person A and B is the same as person B and A).
(Note: if no people in the set is considered invalid, just subtract 1 and get 31).
2006-11-01 07:06:02 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
10 combinations with 2 ppl per combination.... and i think 32 for 0,1,2ppl, 3ppl, 4ppl, and 5ppl
1-2
1-3
1-4
1-5
2-3
2-4
2-5
3-4
3-5
4-5
2006-11-01 07:07:23 · answer #2 · answered by coolpowwow80 3 · 0⤊ 0⤋
Depends on the size of the chain saw!!!
2006-11-01 07:06:59 · answer #3 · answered by tprx899 2 · 1⤊ 0⤋
5 factorial.
2006-11-01 07:06:29 · answer #4 · answered by Terry B 3 · 0⤊ 0⤋