## mathmath333 one year ago Counting question

1. 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}}

2. freckles

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

3. freckles

Or n bulb sockets...

4. mathmath333

5 working bulbs

5. freckles

6. mathmath333

lol my teacher didn't tell any socket thing

7. freckles

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

8. 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

9. freckles

but you can only put in one at a time

10. freckles

since there is only one socket

11. mathmath333

12. freckles

I guess the bulbs come with sockets?

13. freckles

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

14. anonymous

$C_{1}^{5}+C _{2}^{5}+C _{3}^{5}+C _{4}^{5}+C _{5}^{5}=?$

15. mathmath333

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

16. freckles

ok then what @surjithayer says works

17. mathmath333

is it 31

18. mathmath333

and how did u get that

19. ganeshie8

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

20. mathmath333

2^5

21. 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

22. mathmath333

ok thnks

23. ganeshie8

maybe this example is a good place to review the identity : $\dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots+\dbinom{n}{n}~~=~~2^n$

24. 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

25. 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.

26. mathmath333

thst looks useful formula to save time

27. ganeshie8

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