Limit Breaker
New member
- Joined
- Aug 2, 2020
- Messages
- 2
The problem already has the solutions. My doubt is in the second question. Why not apply the same principle for Question 1 and consider the 3 rooms as a big room?
Problem:
You are renovating your entire apartment and want to repaint the walls of each room. The flat consists of two bedrooms, a kitchen, a living room, a bathroom, a study and a hall, or 7 rooms in total. You have at your disposal several colors of paint: white, yellow, orange, red, purple, blue, green, grey and pink.
How many different ways can you paint the house, assuming…
1) …you paint the bathroom, study and hall in white?
If you paint the bathroom, study and hall in white, you only need to think about the other 4 rooms. Now, this problem can be interpreted several different ways, so let us examine each outcome:
a. We paint the other 4 rooms in 4 different colors. That means we have Variations without repetition, so ? = ?! (?−?)! = 9! 5! = 6 × 7 × 8 × 9 = 3,024.
b. We paint the other 4 rooms in 4 different colors. That means we have Variations without repetition. Additionally, we have already used white, so we are down to only 8 colors. Thus, ? = ?! (?−?)! = 8! 4! = 5 × 6 × 7 × 8 = 1,680.
c. We have no restriction on the colors we plan to use in the remaining 4 rooms.
Therefore, we have Variations with repetition, and we can use “white”. Thus, ?̅ = ? ? = 9 4 = 6,561. We phrased the question with the idea of going for interpretation “b”, but we see merit in the other approaches as well.
2) …you paint the two bedrooms in identical color?
If we paint the two bedrooms in the same color, we can think of them as a single big room. Thus, the number of rooms becomes 6 instead of 7.
a. There is no restriction on whether we can repeat any of the colors we use, so we have variations with repetition once more ?̅ = ? ? = 9 6 = 531,441.
b. Alternatively, it is not clearly stated we can repeat values, so let us examine the alternative. If we cannot repeat values, we have variations without repetition, so the formula we use becomes: ? = ?! (?−?)! = 9! 3! = 4 × 5 × 6 × 7 × 8 × 9 = 60,480.
Problem:
You are renovating your entire apartment and want to repaint the walls of each room. The flat consists of two bedrooms, a kitchen, a living room, a bathroom, a study and a hall, or 7 rooms in total. You have at your disposal several colors of paint: white, yellow, orange, red, purple, blue, green, grey and pink.
How many different ways can you paint the house, assuming…
1) …you paint the bathroom, study and hall in white?
If you paint the bathroom, study and hall in white, you only need to think about the other 4 rooms. Now, this problem can be interpreted several different ways, so let us examine each outcome:
a. We paint the other 4 rooms in 4 different colors. That means we have Variations without repetition, so ? = ?! (?−?)! = 9! 5! = 6 × 7 × 8 × 9 = 3,024.
b. We paint the other 4 rooms in 4 different colors. That means we have Variations without repetition. Additionally, we have already used white, so we are down to only 8 colors. Thus, ? = ?! (?−?)! = 8! 4! = 5 × 6 × 7 × 8 = 1,680.
c. We have no restriction on the colors we plan to use in the remaining 4 rooms.
Therefore, we have Variations with repetition, and we can use “white”. Thus, ?̅ = ? ? = 9 4 = 6,561. We phrased the question with the idea of going for interpretation “b”, but we see merit in the other approaches as well.
2) …you paint the two bedrooms in identical color?
If we paint the two bedrooms in the same color, we can think of them as a single big room. Thus, the number of rooms becomes 6 instead of 7.
a. There is no restriction on whether we can repeat any of the colors we use, so we have variations with repetition once more ?̅ = ? ? = 9 6 = 531,441.
b. Alternatively, it is not clearly stated we can repeat values, so let us examine the alternative. If we cannot repeat values, we have variations without repetition, so the formula we use becomes: ? = ?! (?−?)! = 9! 3! = 4 × 5 × 6 × 7 × 8 × 9 = 60,480.
