Question

Counting question

Answer 1

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

Answer 2

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

Answer 3

Or n bulb sockets...

Answer 4

5 working bulbs

Answer 5

Ok what about the sockets.

Answer 6

lol my teacher didn't tell any socket thing

Answer 7

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

Answer 8

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

Answer 9

but you can only put in one at a time

Answer 10

since there is only one socket

Answer 11

answer given =31

Answer 12

I guess the bulbs come with sockets?

Answer 13

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

Answer 14

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

Answer 15

assume u have electricity on all 5 bulbs

Answer 16

ok then what @surjithayer says works

Answer 17

is it 31

Answer 18

and how did u get that

Answer 19

consider a 5 letter word using two different letters

how many total words can you make if repetition is allowed ?

Answer 20

2^5

Answer 21

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

Answer 22

ok thnks

Answer 23

maybe this example is a good place to review the identity :

\[\dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots+\dbinom{n}{n}~~=~~2^n\]

Answer 24

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

Answer 25

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

Answer 26

thst looks useful formula to save time

Answer 27

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