OpenStudy (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.

3 years ago
OpenStudy (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}}

3 years ago
OpenStudy (mathmath333):

I dnt think this is stars ans bas problem

3 years ago
OpenStudy (mathmath333):

stars and bars are used to find integer solutions of the equation eg. a+b+c=x

3 years ago
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 :)

3 years ago
OpenStudy (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

3 years ago
imqwerty (imqwerty):

yea sry that ws not 15*14*13*11*10 i forgot the 12 it wuld be 15*14*13*12*11

3 years ago
OpenStudy (anonymous):

the question is a bit ambiguous . is it asking how many 'different' ways of selecting 5 balls there are

3 years ago
OpenStudy (anonymous):

and we have indistinguishable balls

3 years ago
OpenStudy (mathmath333):

hey same colored balls are identical

3 years ago
OpenStudy (mathmath333):

answer given is 71

3 years ago
OpenStudy (anonymous):

thanks, i will work with this

3 years ago
OpenStudy (mathmath333):

15*14*13*12*11 this works when all 15 balls are distinct

3 years ago
imqwerty (imqwerty):

yea :)

3 years ago
OpenStudy (anonymous):

did you get 71 imquert?

3 years ago
imqwerty (imqwerty):

nope m still wrkin on it :)

3 years ago
OpenStudy (mathmath333):

i m frutated by this que from 2 days

3 years ago
imqwerty (imqwerty):

:D lets finish off the frustration today :)

3 years ago
imqwerty (imqwerty):

hehe see my mod powers :) - http://prntscr.com/8frqsl

3 years ago
OpenStudy (mathmath333):

it happns to every body

3 years ago
OpenStudy (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?

3 years ago
OpenStudy (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

3 years ago
OpenStudy (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.

3 years ago
OpenStudy (anonymous):

'5 + 0 + 0 + 0 + 0 ' means 5 red + 0 black + 0 blue + 0 yellow + 0 green

3 years ago
imqwerty (imqwerty):

:D nice question btw

3 years ago
OpenStudy (mathmath333):

Their should be some trick in this question as it is to be solved in 1-2 min

3 years ago
OpenStudy (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

3 years ago
OpenStudy (mathmath333):

is 7 the answer

3 years ago
OpenStudy (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.

3 years ago
OpenStudy (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)

3 years ago
OpenStudy (mathmath333):

am i amazed ,how did u use the brute force

3 years ago
OpenStudy (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.

3 years ago
OpenStudy (mathmath333):

ok

3 years ago
OpenStudy (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.

3 years ago