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!!!

Hello, digitalis77!

This is a tricky (but immensely satisfying) problem.

Quote:

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),
. . 3 shades of green (G1,G2,G3),
. . 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.

First, select the 5 rooms to be blue and the 2 to be green..
. . There are: . ${7\choose5,2} \:=\:21$ choices.

For each of the 5 blue rooms, there are 4 choices of blue.
. . There are: . $4^5\,=\,1,024$ choices.

For each of 2 green rooms, there are 3 choices of green.
. . There are: . $3^2\,=\,9$ choices.

Therefore, there arte: . $21 \times 1,024 \times 9 \:=\:\boxed{193,536\text{ ways}}$

Quote:

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?

First, select the 3 rooms to be red, 3 rooms to be blue, and 1 to be green.
. . There are: . ${7,\choose3,3,1} \:=\:140$ choices.

For each of the 3 red rooms, there are 2 choices of red.
. . There are: . $2^3 \,=\, 8$ choices.

For each of 3 blue rooms, there are 4 choices of blue.
. . There are: . $4^3 \,=\,64$ choices.

For the one green room, there are 3 choices of green: . $3$ choices.

Therefore, there are: . $140 \times 8 \times 64 \times 3 \:=\:\boxed{215,040\text{ ways}}$

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c) How many ways can the rooms be painted?

For each of the seven rooms, there are nine choices of colors.
. . There are: . $9^7 \:=\:\boxed{4,782,969\text{ ways}}$

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

For each of the 7 rooms, there is a choice of 2 reds.
. . There are: . $2^7 \,=\,\boxed{128\text{ ways}}$

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

The opposite of "at least one green or blue" is "no green or blue" ... all red.

From part (c), there are: . $4,782,969$ ways to paint the rooms
. . of which (d) 128 ways are all red.

Therefore, there are: . $4,782,060 - 128 \:=\:\boxed{4,782,841\text{ ways}}$
. . in which there is at least one green or one blue.

Very helpful

Thank you so much. I had been working on that problem for about 2 hours and just wasn't getting it I completely understand now. Thank you so much!!!!(Clapping)