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Permutations/Combinations

oryxncrake

New member
Joined
Oct 8, 2009
Messages
7
hi

i don't know where else to post this but i need some help on a textbook question i was given about combinatorics.

there is a game that involves a murderer, a mansion, and the weapon used to commit the murder. there are 5 characters in all, 7 rooms in the mansion, and 3 weapons.

the question is.. how many kinds of guesses can i make from choosing one character, one mansion room, and one weapon?

also, is this question a permutation or combination?
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
By the multiplication principle. 5*7*3=105
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588

But I'm certain Miss Scarlet did it in the ballroom with a candlestick.

 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,158
soroban said:


… Miss Scarlet did it in the ballroom with a candlestick.



Soroban, pleeeeze. This is a family-friendly web site.

 

chrisr

Full Member
Joined
Nov 29, 2009
Messages
355
\(\displaystyle Any\ character\ can\ kill\ any\ of\ the\ 4\ other\ characters,\)
\(\displaystyle that's\ 20\ possible\ murders\ if\ they\ are\ all\ potential\ murderers\ and\ victims.\)

\(\displaystyle Using\ any\ of\ the\ 3\ weapons,\ all\ deadly\ in\ some\ way,\)
\(\displaystyle that's\ 20(3)=60\ possible\ ways\ to\ commit\ 20\ killings,\)

\(\displaystyle With\ 7\ choices\ of\ location,\ that's\ 60(7)=420.\)
[sup:3kum0bnp]5[/sup:3kum0bnp]P[sub:3kum0bnp]2[/sub:3kum0bnp][sup:3kum0bnp]3[/sup:3kum0bnp]C[sub:3kum0bnp]1[/sub:3kum0bnp][sup:3kum0bnp]7[/sup:3kum0bnp]C[sub:3kum0bnp]1[/sub:3kum0bnp]=20(3)(7)=420.

\(\displaystyle Should\ we\ include\ self-termination?\)
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, everyone!

If this is the game of CLUE, it already has a victim, Mr. Boddy.

The seven guests (Col. Mustard, Mrs. White, Professor Plum, etc.) are the suspects.

 
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