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.

- mathmath333

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- 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

I dnt think this is stars ans bas problem

- mathmath333

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

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## More answers

- 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

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

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

- anonymous

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

- anonymous

and we have indistinguishable balls

- mathmath333

hey same colored balls are identical

- mathmath333

answer given is 71

- anonymous

thanks, i will work with this

- mathmath333

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

- imqwerty

yea :)

- anonymous

did you get 71 imquert?

- imqwerty

nope m still wrkin on it :)

- mathmath333

i m frutated by this que from 2 days

- imqwerty

:D lets finish off the frustration today :)

- imqwerty

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

- mathmath333

it happns to every body

- 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

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

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

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

- imqwerty

:D nice question btw

- mathmath333

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

- 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

is 7 the answer

- 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

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

am i amazed ,how did u use the brute force

- 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

ok

- 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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