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You have a row of 10 doors (maybe in a wing of a hotel). You are going to paint the doors red and blue, but you don't want two red doors next to each other, (having blue doors next to each other is fine). How many ways can you pain the row of doors?

 

Find the sum: 75 + 77 + 79 + ... 229 + 231.

(This is all the odd numbers from 75 to 231; if you really understand the first two problems, this will be easy).

Guest Jul 5, 2018
 #1
avatar+20179 
0

You have a row of 10 doors (maybe in a wing of a hotel). 

You are going to paint the doors red and blue, but you don't want two red doors next to each other,

(having blue doors next to each other is fine).

How many ways can you pain the row of doors?

 

\(\begin{array}{|r|r|} \hline \text{red doors in the row} & \text{no two red doors next to each other} \\ \hline 0 & 1 \\ 1 & 10 \\ 2 & 36 \\ 3 & 56 \\ 4 & 35 \\ 5 & 6 \\ \hline \text{sum} & 144 \\ \hline \end{array}\)

 

You can pain the row of doors 144 ways.

 

laugh

heureka  Jul 6, 2018
 #2
avatar+20179 
0

Find the sum: 75 + 77 + 79 + ... 229 + 231.

 

\(\begin{array}{|rcll|} \hline && 75 + 77 + 79 + \cdots 229 + 231 \\ &=& (75+0) + (75+2) + (75+4) + \ldots + (75+154) + (75+156) \\ &=& (75+0) + (75+1\cdot2) + (75+2\cdot2) + \ldots + (75+77\cdot2) + (75+78\cdot2) \\ &=& 75 + 75 \cdot 78 + 2 \cdot (1+2+\ldots + 77+78) \\ &=& 75 \cdot 79 + 2 \cdot \dfrac{(1+78)\cdot78}{2} \\ &=& 75 \cdot 79 + 2 \cdot \dfrac{79 \cdot 78}{2} \\ &=& 75 \cdot 79 + 79 \cdot 78 \\ &=& 79\cdot(75+78) \\ &=& 79\cdot 153 \\ &\mathbf{=}& \mathbf{12087} \\ \hline \end{array}\)

 

laugh

heureka  Jul 6, 2018

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