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mathmath333

  • one year ago

An urn has 5 red balls ,4 black balls,3 blue balls,2 yellow balls and 1 green ball. In how many ways can 5 balls be selected.

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{An urn has 5 red balls ,4 black balls,3 blue balls,2 yellow balls}\hspace{.33em}\\~\\ & \normalsize \text{and 1 green ball.}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can 5 balls be selected.}\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    I dnt think this is stars ans bas problem

  3. mathmath333
    • one year ago
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    stars and bars are used to find integer solutions of the equation eg. a+b+c=x

  4. imqwerty
    • one year ago
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    well there are no other conditions given like there shuld be atleast 2 yellow ones or something like that so we can consider like u have a total of 15balls nd u need to find out number of ways to select 5balls so it wuld be 15*14*13*11*10 but wait we can have cases like this-|dw:1442129915453:dw| so we are getting similar cases :) nd we need to fix them but how :)

  5. anonymous
    • one year ago
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    you can write out explicitly the options r = red, b = black , B = blue, y = yellow, g = green r1, r2, r3 , r4, r5 b1, b2, b3 , b4, B1, B2, B3 y1, y2 g there are 5+4+3+2+1 = 15 balls there are 15*14*13*12*11 ways to choose 5 balls. I am assuming order counts because you can select r1,r2,r3,r4,r5 in different ways

  6. imqwerty
    • one year ago
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    yea sry that ws not 15*14*13*11*10 i forgot the 12 it wuld be 15*14*13*12*11

  7. anonymous
    • one year ago
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    the question is a bit ambiguous . is it asking how many 'different' ways of selecting 5 balls there are

  8. anonymous
    • one year ago
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    and we have indistinguishable balls

  9. mathmath333
    • one year ago
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    hey same colored balls are identical

  10. mathmath333
    • one year ago
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    answer given is 71

  11. anonymous
    • one year ago
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    thanks, i will work with this

  12. mathmath333
    • one year ago
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    15*14*13*12*11 this works when all 15 balls are distinct

  13. imqwerty
    • one year ago
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    yea :)

  14. anonymous
    • one year ago
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    did you get 71 imquert?

  15. imqwerty
    • one year ago
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    nope m still wrkin on it :)

  16. mathmath333
    • one year ago
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    i m frutated by this que from 2 days

  17. imqwerty
    • one year ago
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    :D lets finish off the frustration today :)

  18. imqwerty
    • one year ago
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    hehe see my mod powers :) - http://prntscr.com/8frqsl

  19. mathmath333
    • one year ago
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    it happns to every body

  20. anonymous
    • one year ago
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    another way to look at this is to list the possible number of colored balls Red = 0,1,2,3,4,5 Black = 0,1,2,3,4 Blue = 0,1,2,3 yellow = 0,1,2 green = 0,1 how many ways can you add up to 5?

  21. anonymous
    • one year ago
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    starting with Red = 5, we have 5 + 0 + 0 + 0 + 0 4 + 1 + 0 + 0 + 0 4 + 0 + 1 + 0 + 0 4 + 0 + 0 + 1 + 0 4 + 0 + 0 + 0 + 1

  22. anonymous
    • one year ago
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    then 3 + 2 + 0 + 0 + 0 3 + 0 + 2 + 0 + 0 3 + 0 + 0 + 2 + 0 then list with 3 reds. This is brute force method and long.

  23. anonymous
    • one year ago
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    '5 + 0 + 0 + 0 + 0 ' means 5 red + 0 black + 0 blue + 0 yellow + 0 green

  24. imqwerty
    • one year ago
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    :D nice question btw

  25. mathmath333
    • one year ago
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    Their should be some trick in this question as it is to be solved in 1-2 min

  26. anonymous
    • one year ago
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    there are 7 ways to add up to 5 The seven partitions of 5 are: 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 now count how many ways you can fit these in the boxes, with the restriction that first ball is between 0 and 5 second ball is between 0 and 4, etc

  27. mathmath333
    • one year ago
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    is 7 the answer

  28. mathmate
    • one year ago
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    The answer is 71 by enumeration, but I have not found a systematical way to do it. If you want to see the list, let me know.

  29. mathmate
    • one year ago
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    Here's the list of the combinations, first number is the green [0,1], last number is red [0,5]. Perhaps someone can figure out the logic. (0, 0, 0, 0, 5) (0, 0, 0, 1, 4) (0, 0, 0, 2, 3) (0, 0, 0, 3, 2) (0, 0, 0, 4, 1) (0, 0, 1, 0, 4) (0, 0, 1, 1, 3) (0, 0, 1, 2, 2) (0, 0, 1, 3, 1) (0, 0, 1, 4, 0) (0, 0, 2, 0, 3) (0, 0, 2, 1, 2) (0, 0, 2, 2, 1) (0, 0, 2, 3, 0) (0, 0, 3, 0, 2) (0, 0, 3, 1, 1) (0, 0, 3, 2, 0) (0, 1, 0, 0, 4) (0, 1, 0, 1, 3) (0, 1, 0, 2, 2) (0, 1, 0, 3, 1) (0, 1, 0, 4, 0) (0, 1, 1, 0, 3) (0, 1, 1, 1, 2) (0, 1, 1, 2, 1) (0, 1, 1, 3, 0) (0, 1, 2, 0, 2) (0, 1, 2, 1, 1) (0, 1, 2, 2, 0) (0, 1, 3, 0, 1) (0, 1, 3, 1, 0) (0, 2, 0, 0, 3) (0, 2, 0, 1, 2) (0, 2, 0, 2, 1) (0, 2, 0, 3, 0) (0, 2, 1, 0, 2) (0, 2, 1, 1, 1) (0, 2, 1, 2, 0) (0, 2, 2, 0, 1) (0, 2, 2, 1, 0) (0, 2, 3, 0, 0) (1, 0, 0, 0, 4) (1, 0, 0, 1, 3) (1, 0, 0, 2, 2) (1, 0, 0, 3, 1) (1, 0, 0, 4, 0) (1, 0, 1, 0, 3) (1, 0, 1, 1, 2) (1, 0, 1, 2, 1) (1, 0, 1, 3, 0) (1, 0, 2, 0, 2) (1, 0, 2, 1, 1) (1, 0, 2, 2, 0) (1, 0, 3, 0, 1) (1, 0, 3, 1, 0) (1, 1, 0, 0, 3) (1, 1, 0, 1, 2) (1, 1, 0, 2, 1) (1, 1, 0, 3, 0) (1, 1, 1, 0, 2) (1, 1, 1, 1, 1) (1, 1, 1, 2, 0) (1, 1, 2, 0, 1) (1, 1, 2, 1, 0) (1, 1, 3, 0, 0) (1, 2, 0, 0, 2) (1, 2, 0, 1, 1) (1, 2, 0, 2, 0) (1, 2, 1, 0, 1) (1, 2, 1, 1, 0) (1, 2, 2, 0, 0) ('count=', 71)

  30. mathmath333
    • one year ago
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    am i amazed ,how did u use the brute force

  31. mathmate
    • one year ago
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    When I cannot solve a problem, brute force is required, and the computer is my best friend for that purpose! lol If you use Python, I can send you the code.

  32. mathmath333
    • one year ago
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    ok

  33. mathmate
    • one year ago
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    Table of number of ways to obtain total no. of balls progressively sum of number of balls 0 1 2 3 4 5 red (5) 1 1 1 1 1 1 black(4) 1 1 1 1 1 (5+4) 1 2 3 4 5 5 blue(3) 1 1 1 1 (5+4+3) 1 3 6 10 14 17 yellow(2) 1 1 1 (5...2) 1 4 10 19 30 41 green(1) 1 1 (5...1) 1 5 14 29 49 71 Number of ways to make the given sum (0 - 5) for given colour combinations. Explanations: The first line of the table enumerates the ways to pick the red balls to get a given total of one to five balls. The next line is similar, but for black balls. The third line is the number of ways to choose x balls (0<=x<=5) when red and black are combined. After that, we add progressively blue, yellow and green. The last line is when there are all 15 balls, with 71 ways to get a total of five balls. Following are some examples: for 5 reds and 4 blacks, there are 1*1=1 way to get a sum of 0 ball 1*1+1*1=2 ways to get 1 ball, ... 1*1+1*1+1*1+1*1+1*1=5 ways to get 4 balls 1*1+1*1+1*1+1*1+1*1=5 ways to get 5 balls The result is shown on the line (5+4) For 5 reds, 4 blacks, and 3 blues, tag on the blues below the line (5+4) and work similarly. 1*1=1 way to get a sum of zero ball. 2*1+1*1=3 ways to get 1 ball, 3*1+2*1+1*1=6 ways to get 2 balls, ... 3*1+4*1+5*1+5*1=17 ways to get a sum of 5 balls. Continuing this way, we have, with all 15 balls, 71 ways to get a sum of exactly 5 balls.

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