Combinatorics

• Apr 15th 2008, 07:07 AM
Tom101
Combinatorics
Supopose tha ti have 10 computer games that i need to play . Only one game me be played at any time
in how many ways can the games be run if...

a)Four of the games are considered better the others and player first
b) there is no restriction on the playing ordder of the games
c) the games can be sepertaed in to order into..three in top prioirty ( to played first) five in medium ( to be played straight after the top three) five low to be player last

many thanks
tom
• Apr 15th 2008, 03:22 PM
Soroban
Hello, Tom!

Quote:

Suppose that I have 10 computer games that i need to play.
Only one game may be played at any time.

In how many ways can the games be run if:

a) Four of the games are considered better the others and played first

The first four can be ordered in \$\displaystyle 4!\$ ways.
The last six can be ordered in \$\displaystyle 6!\$ ways.

Hence, there are: .\$\displaystyle (4!)(6!) \:=\:(24)(720) \:=\:2,880\$ ways.

Quote:

b) There is no restriction on the playing order of the games
There are: .\$\displaystyle 10! \:=\:3,628,800\$ ways.

Quote:

c) The games can be separated into order into:
three in top priority (played first),
five in medium (played straight after the top three),
five in low (played last). . . . . 13 games?

The top priority games can be ordered in \$\displaystyle 3!\$ ways,
the medium priority games can be ordered in \$\displaystyle 5!\$ ways,
the low priority games can be ordered in \$\displaystyle 5!\$ ways.

Hence, there are: .\$\displaystyle (3!)(5!)(5!) \:=\:(6)(120)(120) \;=\;86,400\$ ways.