- mathmath333

Counting question

- katieb

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- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{Their are 5 bulbs in a room.}\hspace{.33em}\\~\\
& \normalsize \text{Find the number of ways in which the room can be lightened}\hspace{.33em}\\~\\
\end{align}}\)

- freckles

Are there 5 bulb sockets or 1 bulb socket? I don't know the names of them...

- freckles

Or n bulb sockets...

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## More answers

- mathmath333

5 working bulbs

- freckles

Ok what about the sockets.

- mathmath333

lol my teacher didn't tell any socket thing

- freckles

well if there is 1 socket then there are only 5 ways to light the room

- freckles

You 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

but you can only put in one at a time

- freckles

since there is only one socket

- mathmath333

answer given =31

- freckles

I guess the bulbs come with sockets?

- freckles

Like I just don't see how you can light a room without the bulb having electricity pumped into it.

- anonymous

\[C_{1}^{5}+C _{2}^{5}+C _{3}^{5}+C _{4}^{5}+C _{5}^{5}=?\]

- mathmath333

assume u have electricity on all 5 bulbs|dw:1440345582104:dw|

- freckles

ok then what @surjithayer says works

- mathmath333

is it 31

- mathmath333

and how did u get that

- ganeshie8

consider a 5 letter word using two different letters
how many total words can you make if repetition is allowed ?

- mathmath333

2^5

- freckles

A, 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

- mathmath333

ok thnks

- ganeshie8

maybe 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

Treat 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

\[C _{1}^{5}=\frac{ 5 }{ 1 }=5\]
\[C _{2}^{5}=\frac{ 5\times 4 }{ 2 \times1 }=10\]
\[C _{3}^{5}=\frac{ 5 \times 4 \times 3 }{ 3 \times 2 \times 1 }=10\]
\[C _{4}^{5}=\frac{ 5*4*3*2 }{ 4*3*2*1 }=5\]
\[C _{5}^{5}=\frac{ 5*4*3*2*1 }{ 5*4*3*2*1 }=1\]
Add them all.

- mathmath333

thst looks useful formula to save time

- ganeshie8

do not memorize the formula, study the pascal triangle instead
you will automatically remember that identity and other cool stuff..

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