# Combinations and permutations

• Mar 18th 2008, 12:11 PM
digitalis77
Combinations and permutations
The 7 rooms of a new home are to be painted. Each room is to be painted one color from a selection of 4 shades of blue (B1,B2,B3,B4) and 3 shades of green (G1,G2,G3) and 2 shades of red (R1,R2). More than one room may have the same color.

Count the number of ways each of the following steps can be completed to determine the number of different ways to paint the 7 rooms.

a.) If 5 are to be painted a shade of blue and 2 a shade of green.

b.) 3 rooms some shade of red, 3 rooms some shade of blue and one room some shade of green, how many different ways can the seven rooms be painted?

c.)How many ways can the rooms be painted?

d.) How many different ways can all the rooms be painted red?

e.) How many different ways can the rooms be painted if atleast one room is green or blue?

Any help would be tremendously appreciated!!!(Crying)
• Mar 18th 2008, 01:02 PM
iknowone
For each problem I assume that it is not fixed what rooms will be blue, red or green; just that some number of the total must be a particular color.

a) If 5 are to be painted a shade of blue and 2 a shade of green.

First choose which 5 of the 7 rooms will be blue. There are C(7,5) such choices. Each of the blue rooms can be 1 of 4 different shades. So there are 4 choices for the first room, 4 for the second, and so on. Thus there are 4^5 ways the 5 blue rooms could be painted. Likewise there are 3 ways for the one green room to be painted and 3 for the other green room. So there are C(7,5)(4^5)(3^2) total ways to paint the rooms with the above restrictions.

b) 3 rooms some shade of red, 3 rooms some shade of blue and one room some shade of green, how many different ways can the seven rooms be painted?

In a similar way as the previous problem you can see that there are C(7,3) choices of the 3 red rooms, C(4,3) choices left for the 3 blue rooms, and the last room (green) is forced by these first two choices. Then there are C(7,3)C(4,3)(2^3)(4^3)(3) ways to paint the 7 rooms with the above restrictions.

c) How many ways can the rooms be painted?

In this problem you can think of each room individually as having 9 different choices of color. (4 blue + 3 green + 2 red). So there are 9^7 ways to paint all 7 rooms without restriction.

d) all the rooms are painted red?

2 red shade choices for each room. So the total is 2^7.

e) at least one room is green or blue?

Ask it in a different way. How many ways are there to paint the rooms without any blue and without any green.
• Mar 18th 2008, 01:26 PM
topsquark
Quote:

Originally Posted by digitalis77
The 7 rooms of a new home are to be painted. Each room is to be painted one color from a selection of 4 shades of blue (B1,B2,B3,B4) and 3 shades of green (G1,G2,G3) and 2 shades of red (R1,R2). More than one room may have the same color.

Count the number of ways each of the following steps can be completed to determine the number of different ways to paint the 7 rooms.

a.) If 5 are to be painted a shade of blue and 2 a shade of green.

b.) 3 rooms some shade of red, 3 rooms some shade of blue and one room some shade of green, how many different ways can the seven rooms be painted?

c.)How many ways can the rooms be painted?

d.) How many different ways can all the rooms be painted red?

e.) How many different ways can the rooms be painted if atleast one room is green or blue?

Any help would be tremendously appreciated!!!(Crying)