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mathmath333
 one year ago
Counting question
mathmath333
 one year ago
Counting question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} & \normalsize \text{Their are 5 bulbs in a room.}\hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which the room can be lightened}\hspace{.33em}\\~\\ \end{align}}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Are there 5 bulb sockets or 1 bulb socket? I don't know the names of them...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Ok what about the sockets.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2lol my teacher didn't tell any socket thing

freckles
 one year ago
Best ResponseYou've already chosen the best response.3well if there is 1 socket then there are only 5 ways to light the room

freckles
 one year ago
Best ResponseYou've already chosen the best response.3You know you can put bulb A in the socket you can put bulb B in the socket bulb C in the socket bulb D in the socket bulb E in the socket

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but you can only put in one at a time

freckles
 one year ago
Best ResponseYou've already chosen the best response.3since there is only one socket

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I guess the bulbs come with sockets?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Like I just don't see how you can light a room without the bulb having electricity pumped into it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[C_{1}^{5}+C _{2}^{5}+C _{3}^{5}+C _{4}^{5}+C _{5}^{5}=?\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2assume u have electricity on all 5 bulbsdw:1440345582104:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok then what @surjithayer says works

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2and how did u get that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1consider a 5 letter word using two different letters how many total words can you make if repetition is allowed ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3A, B, C, D, E all of these can be turned on... 1 way 4 of these can be turned on... 5 choose 4 3 of these can be turned on...5 choose 3 2 of these can be turned on 5 choose 2 1 of these can be turned on...5 choose 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1maybe this example is a good place to review the identity : \[\dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots+\dbinom{n}{n}~~=~~2^n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Treat the problem as counting the number of binary strings {0,1} of length 5 On = 1, Off = 0 Example: 10001 Each light has two ways, it can be on or off. By multiplication principle there are 2x2x2x2x2 = 2^5 ways to turn on and off the lights. two ways for first light, and two ways for second light, ... 2 ways for fifth light Since you want at least one light on, we ignore 00000 2^5 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[C _{1}^{5}=\frac{ 5 }{ 1 }=5\] \[C _{2}^{5}=\frac{ 5\times 4 }{ 2 \times1 }=10\] \[C _{3}^{5}=\frac{ 5 \times 4 \times 3 }{ 3 \times 2 \times 1 }=10\] \[C _{4}^{5}=\frac{ 5*4*3*2 }{ 4*3*2*1 }=5\] \[C _{5}^{5}=\frac{ 5*4*3*2*1 }{ 5*4*3*2*1 }=1\] Add them all.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2thst looks useful formula to save time

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1do not memorize the formula, study the pascal triangle instead you will automatically remember that identity and other cool stuff..
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