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mathmath333

  • one year ago

Counting question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Their are 5 bulbs in a room.}\hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which the room can be lightened}\hspace{.33em}\\~\\ \end{align}}\)

  2. freckles
    • one year ago
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    Are there 5 bulb sockets or 1 bulb socket? I don't know the names of them...

  3. freckles
    • one year ago
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    Or n bulb sockets...

  4. mathmath333
    • one year ago
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    5 working bulbs

  5. freckles
    • one year ago
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    Ok what about the sockets.

  6. mathmath333
    • one year ago
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    lol my teacher didn't tell any socket thing

  7. freckles
    • one year ago
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    well if there is 1 socket then there are only 5 ways to light the room

  8. freckles
    • one year ago
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    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
    • one year ago
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    but you can only put in one at a time

  10. freckles
    • one year ago
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    since there is only one socket

  11. mathmath333
    • one year ago
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    answer given =31

  12. freckles
    • one year ago
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    I guess the bulbs come with sockets?

  13. freckles
    • one year ago
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    Like I just don't see how you can light a room without the bulb having electricity pumped into it.

  14. anonymous
    • one year ago
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    \[C_{1}^{5}+C _{2}^{5}+C _{3}^{5}+C _{4}^{5}+C _{5}^{5}=?\]

  15. mathmath333
    • one year ago
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    assume u have electricity on all 5 bulbs|dw:1440345582104:dw|

  16. freckles
    • one year ago
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    ok then what @surjithayer says works

  17. mathmath333
    • one year ago
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    is it 31

  18. mathmath333
    • one year ago
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    and how did u get that

  19. ganeshie8
    • one year ago
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    consider a 5 letter word using two different letters how many total words can you make if repetition is allowed ?

  20. mathmath333
    • one year ago
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    2^5

  21. freckles
    • one year ago
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    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
    • one year ago
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    ok thnks

  23. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[C _{1}^{5}=\frac{ 5 }{ 1 }=5\] \[C _{2}^{5}=\frac{ 5\times 4 }{ 2 \times1 }=10\] \[C _{3}^{5}=\frac{ 5 \times 4 \times 3 }{ 3 \times 2 \times 1 }=10\] \[C _{4}^{5}=\frac{ 5*4*3*2 }{ 4*3*2*1 }=5\] \[C _{5}^{5}=\frac{ 5*4*3*2*1 }{ 5*4*3*2*1 }=1\] Add them all.

  26. mathmath333
    • one year ago
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    thst looks useful formula to save time

  27. ganeshie8
    • one year ago
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    do not memorize the formula, study the pascal triangle instead you will automatically remember that identity and other cool stuff..

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