Math Problem

[Ronn_Bman]

Ok, this should be a simple one based on the problems I've seen resolved here, but simple or not, I don't remember the formula.

You have four levers, each can be set in two positions, what is the total number of unique lever configurations?

I was thinking (4x2)+(3x2)+(2x2)+(2x1)+(2x0), but that equals 20, and I can only come up with 16 configurations?

Don't laugh! I've been out of school longer than I was in school. [img]smile.gif[/img]

[andrewas]

2^4, or 16.

Each lever has 2 positons. With two levers there are 2*2 positions. With three its 2*2*2, and with four its simpler just to say 2^4.

Originally posted by andrewas: 2^4, or 16. Each lever has 2 positons. With two levers there are 2*2 positions. With three its 2*2*2, and with four its simpler just to say 2^4. or 4^2 [img]smile.gif[/img]

[andrewas]

Originally posted by MagiK: quote: Originally posted by andrewas: 2^4, or 16. Each lever has 2 positons. With two levers there are 2*2 positions. With three its 2*2*2, and with four its simpler just to say 2^4. or 4^2 [img]smile.gif[/img]

[/QUOTE]Thats an exception. It works with 4 levers. But with 5, 2^5 is 32 and 5^2 is 25. Simpler to stick with the general case.

[Ronn_Bman]

Thanks Guys!

Now the bonus question....

What the crap is that formula I listed....lol. [img]smile.gif[/img]

Originally posted by Ronn_Bman: Thanks Guys! Now the bonus question.... What the crap is that formula I listed....lol. [img]smile.gif[/img]

Beats me BMan [img]smile.gif[/img] but congrats to going back, glad to know Im not the only senior student here [img]smile.gif[/img]

Andrew! Picky, Picky, Picky hehehe

[andrewas]

Originally posted by Ronn_Bman: Thanks Guys! Now the bonus question.... What the crap is that formula I listed....lol. [img]smile.gif[/img]

Good question. 2*n!. It does resemble some not entirely unrealated forumlae though, it gives the number of orderings, order does matter, from a set. OK, Its vaguely related at best.

[whacky]

Originally posted by andrewas: quote: Originally posted by Ronn_Bman: Thanks Guys! Now the bonus question.... What the crap is that formula I listed....lol. [img]smile.gif[/img]

Good question. 2*n!. It does resemble some not entirely unrealated forumlae though, it gives the number of orderings, order does matter, from a set. OK, Its vaguely related at best.[/QUOTE]Dang andrew you beat me to it correct me if i'm wrong though, the formula 2^n gives the number of possible sets not including a full set and an empty set ??