mathmath333
  • mathmath333
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{An urn has 5 red balls ,4 black balls,3 blue balls,2 yellow balls}\hspace{.33em}\\~\\ & \normalsize \text{and 1 green ball.}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can 5 balls be selected.}\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
I dnt think this is stars ans bas problem
mathmath333
  • mathmath333
stars and bars are used to find integer solutions of the equation eg. a+b+c=x

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imqwerty
  • imqwerty
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 :)
anonymous
  • anonymous
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
imqwerty
  • imqwerty
yea sry that ws not 15*14*13*11*10 i forgot the 12 it wuld be 15*14*13*12*11
anonymous
  • anonymous
the question is a bit ambiguous . is it asking how many 'different' ways of selecting 5 balls there are
anonymous
  • anonymous
and we have indistinguishable balls
mathmath333
  • mathmath333
hey same colored balls are identical
mathmath333
  • mathmath333
answer given is 71
anonymous
  • anonymous
thanks, i will work with this
mathmath333
  • mathmath333
15*14*13*12*11 this works when all 15 balls are distinct
imqwerty
  • imqwerty
yea :)
anonymous
  • anonymous
did you get 71 imquert?
imqwerty
  • imqwerty
nope m still wrkin on it :)
mathmath333
  • mathmath333
i m frutated by this que from 2 days
imqwerty
  • imqwerty
:D lets finish off the frustration today :)
imqwerty
  • imqwerty
hehe see my mod powers :) - http://prntscr.com/8frqsl
mathmath333
  • mathmath333
it happns to every body
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
'5 + 0 + 0 + 0 + 0 ' means 5 red + 0 black + 0 blue + 0 yellow + 0 green
imqwerty
  • imqwerty
:D nice question btw
mathmath333
  • mathmath333
Their should be some trick in this question as it is to be solved in 1-2 min
anonymous
  • anonymous
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
mathmath333
  • mathmath333
is 7 the answer
mathmate
  • mathmate
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.
mathmate
  • mathmate
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)
mathmath333
  • mathmath333
am i amazed ,how did u use the brute force
mathmate
  • mathmate
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.
mathmath333
  • mathmath333
ok
mathmate
  • mathmate
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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